HIGHER ARITHMETIC, DESIGNED FOR THE USE OF HIGH SCHOOLS, ACADEMIES, AND COLLEGES; IN WHICH SOME ENTIRELY NEW PRINCIPLES ARE DEVELOPED AND MANY CONCISE AND EASY RULES GIVEN1 WHICH HAVE NEVER BEFORE APPEARED IN ANY ARITHMETIC: VWITH AN APPENDIX. BY GEORGE R. PERKINS, A. M., PRINCIPAL AND PROFESSOR OF MATHEMATICS OF THE NORMAL SCHOOL OF THE STATI OF NEW YORK, AUTHOR OF ELEMENTARY ARITHMETIC, ELEMENTARY ALGEBRA, TREATISE ON ALGEBRA, ELEMlENTS OF GEOMETRY, ETC, ETC. STEREOTYPE EDITION, REVISED AND IMPROVED. UTICA: PUBLISHED BY II. H. HAWLEY & CO HARTFORD: J. H. MATHER & CO. 1849. ENTERED according to Act of Conr'ess, in the year 1848, by GEORGE R. PERKINS, In the Clerk's Office of the Northern District of New York. STEREOTYPED BY PRINTED BY RICHARD II. HOBBS, CASE, TIFPIANY & CO., HARTFORD, CONN. HARTFORD, CONN. PREFACE TO THE SECOND EDITION. IN preparing this edition, the whole work has been carefully revised, and many of the subjects re-written and modified, with a view to make them more clearly comprehended by the pupil. Having explained fully, in the ELEMENTARY ARITHMETIC, the subjects of Duodecimals and Alligation, we have omitted them in the present edition. We have excluded the chapter on Permutations, Combinations, and Variations, to make room for what we deemed matter of more importance. We have also placed the chapters on the Progressions after Evolution, because several cases under the Progressions are wrought by the aid of Evolution. Under the first chapter, we have added some new things concerning Prime Numbers, and have explained the use of ERATOSTHENES' SIEVE in finding these numbers. After Continued Fractions, we have added LAMBERT'S Method of De-compounding Vulgar Fractions. We have given a new rule for the Equation of Payments, under the subject of Discount. This rule is deduced from considering the equivalent present values. By this method, we discover that the usual rule is erroneous, if we consider the present value of the several debts. We have added the answers to all the questions; and, where we thought the operations too difficult, we have either given the whole work, or so much as to leave nothing to perplex the beginner. In all these modifications we have been guided, not only by our own practical experience in using the book, but by the many kind suggestions which have been received from distinguished teachers. GEO. R. PERKINS. UTICA, September, 1844. PREFACE TO THE FOURTH EDITION. THE present edition, which is stereotyped, has been carefully revised, and, in some cases, new matter added in the body of the work, without, however, changing the order of the different articles. But what most distinguishes it from the preceding ones, is the addition of an APPENDlx. In this Appendix, we have discussed pretty fully what may properly be considered the philosophy of some of the more difficult operations, as well as interesting properties of numbers. It is believed that much here added will be found instructive and interesting to the lover of arithmetical operations. GEO. R. PERKINS. UTICA, April, 184 CONTENTS. CHAPTER I. Page. DEFINITIONS,............... 9 Symbols,.............10 Examples illustrating definitions and symbols,..... 12 Multiplication of compound expressions,....... 14 Singular property of the figure 9,........ 17 Prime numbers............. 17 Table of prime numbers,. 19 Numbers decomposed into prime factors,...... 21 First method of finding the greatest common measure,,. 23 Second " " " " ". 24 First method of finding the least common multiple,.... 25 Second " " " " "..... 26 To find all the divisors of any number,. 29 The number of divisors given, without exhibiting them,.. 30 CHAPTER II. Vulgar fractions defined,.... 32 Reduction of fractions,..... 33 To reduce simple fractions to their lowest terms,... 34 To reduce improper fractions to mixed numbers,. 36 To reduce mixed numbers to improper fractions,... 37 To reduce compound fractions to simple ones,... 38 To reduce fractions to a common denominator,. 41 To reduce fractions to their least common denominator,... 43 Addition of fractions,.... 44 Subtraction of fractions,...... 46 Multiplication of fractions,..... 47 Division of fractions,....... 49 Reduction of complex fractions,.. 50 Reduction of fractions to a given denominator,.....51 Reduction of denominate fractions defined,.. 53 1* vi CONTENTS. Page. To reduce fractions from one denominate value to another,. 53 To find what fractional part one quantity is of another,.. 55 Reduction of denominate fractions to whole denominate numbers 57 CHAPTER III. Decimal fractions defined,...... 59 Numeration table of whole numbers and decimals,... 59 Addition of decimals,.... 60 Subtraction of decimals,. 62 Multiplication of decimals,....... 63 Abridged multiplication of decimals,..... 64 Division of decimals,...69 Abridged division of decimals,..... 70 To change a vulgar fraction to a decimal,.... 74 Whether a vulgar fraction can be accurately decimated,.. 75 When it can be expressed, to determine the number of decimals 76 Method of decimating when many figures are required,.. 77 Method of decimating by continued multiplication,. 79 When the decimals do not terminate, they recur in periods, 80 Simple repetends,.... 81 Compound repetends,... 81 Perfect repetends,... 82 Singular properties of perfect repetends explained by aid of a circle,..............86 To change a decimal fraction into a vulgar one,... 90 CHAPTER IV. Continued fractions defined,. 94 To convert a vulgar fraction into a continued fraction,.. 96 To convert a continued fraction into a vulgar fraction, 98 Use of continued fractions,. 102 Roots of surds by continued fractions,........ 104 Properties of continued fractions,. 105 Lambert's method of de-compounding vulgar fractions, 106 CHAPTER V. Rule of three,.... 110 Compound proportion,...... 119 CONTENTS. Vil CHAPTER VI. Page. Simple interest defined,.... 126 To find the interest on $1, for any time, at 6 per cent.,.. 126 To find the interest on any sum, for any time, at 6 per cent.,. 128 To find the interest on any sum, for any time, at any per cent., 129 Table of interest at 7 per cent.,... 131 When partial payments are made on notes, bonds, etc.,.. 134 Given the time, rate per cent., and interest, to find the principal, 140 Principal, time, and interest given, to find the per cent.,.. 141 Principal, rate per cent., and interest, to find the time,.. 142 Discount defined,............... 143 To find the discount and present worth,.. 143 Table giving the present worth at 7 per cent.,... 145 New rule for equation of payments,....... 148 CHAPTER VII. Compound interest defined,... 151 Table to facilitate the casting of compound interest,... 153 Compound discount defined,............ 155 To find the present worth of a sum at compound discount,. 156 Table to assist in finding the present worth at compound interest 157 Annuities defined.............. 159 To find the amount of an annuity,......... 159 Table by which to find the amount of an annuity,... 161 To find the present worth of an annuity,. 162 Table by which to find the present worth of an annuity,.. 164 To find the present worth of an annuity in reversion,... 166 Tables-more for curiosity than utility,.. 169 CHAPTER VIII. Banking defined,..... 172 To find the banking discount,.......... 173 Given the present worth, time, and rate, to find the amount,. 173 The same found by the aid of a table,.... 174 The per cent. which banks receive exceeds the legal rate,. 175 Table showing this excess,....... 177 CHAPTER IX. Involution defined,............... 178 CONTENTS. viii Page. Evolution defined,......179 Extraction of the square root of a whole number,.. 180 When the number has decimals,......... 182 When the number-is a vulgar fraction,..... 183 When many figures are required,...... 184 Examples involving the principles of the square root,.. 186 Extraction of the cube root of a whole number,.. 193 When the number has decimals,... 197 When the number is a vulgar fraction,... 198 When many decimals are required,... 199 Examples involving the principles of the cube root, 203 Extraction of roots of all powers,... 205 CHAPTER X. Arithmetical progression,..... 211 CHAPTER XI. Geometrical progression,.. 232 Summation of decreasing infinite series,... 246 CHAPTER XII. Miscellaneous questions solved by analysis,. 248 CHAPTER XIII. Miscellaneous questions,.... 273 APPENDIX,................. 283 ARITHMETIC. CHAPTER I. DEFINITIONS. 1. ANY whole number is called an integer. 2. Any number which can be divided by 2, without a remainder, is called an even number. 3. All numbers which can not be divided by 2, without a remainder, are called odd numbers. 4. Any number which can be produced by multiplying two or more numbers together, each of which is greater than a unit, is called a composite number. Thus, 35 is a composite number, since it can be produced by multiplying 5 and 7 together. 5. The numbers which are multiplied together to produce a composite number, are calledfactors. Thus, 3 and 8 are factors of 24; so, also, are 4 and 6. 6. A composite number which is composed of two equal factors, is called a square number. Thus, 4, 9, 16, and 49, are square numbers. 7. A composite number which is composed of three equal factors, is called a cube number. Thus, 8, 27, and 64, are cube numbers. 8. One of the equal factors which compose a square number, is called the square root of the number. Thus, 7 is called the square root of 49. 9. One of the equal factors which compose a cube 10 HIGHER ARITHMETIC. number, is called the cube root of the number. Thus, 3 is the cube root of 27. 10. All numbers which are not composite, are called prime numbers. Thus, 1, 2, 3, 5, 7, 11, and 13, are prime numbers. 11. Unity divided by a number is the reciprocal of that number. Thus, 2, 3, and 4, are the reciprocals of 2, 3, and 4. 12. A number taken without regard to the particular kind of unit, is called an abstract number. 13. A number considered in reference to a particular unit, is called a concrete number, or a denominate number. Thus, 8 yards is a denominate number, whose unit is one yard; 37 days is a denominate number, whose unit is one day. SY MB 0 L S 2. THE symbol =, is called the sign of equality; and denotes that the quantities between which it is placed are equal or equivalent to each other. Thus, $1 =100 cents; which is read, one dollar equals one hundred cents. 2.. The symbol +, is called plus; and denotes that the quantities between which it is placed are to be added together. Thus, 6 + 2= 8; which is read, six and two added equals eight. S. The symbol -, is called minus, and denotes that the quantity which is placed at the right of it is to be subtracted from the quantity on the left. Thus, 6-24 -4; which is read, six diminished by two equals four. SY il BOLS- 1l 4. The symbol x, is called the sign of multiplication, and denotes that the quantities between which it is placed are to be multiplied together. Thus, 6x 2= 12; which is read, six multiplied by two equals twelve. Multiplication is sometimes expressed by a dot, (.); Thus, 3. 4, is the same as 3 x 4. 5. The symbol -, is called the sign of division, and denotes that the quantity on the left of it is to be divided by the quantity on the right. Thus, 6 2=3; which is read, six divided by two equals three. Division is also denoted by placing the divisor under the dividend, with a horizontal line between them, like a vulgar fraction. Thus, 4 is the same as 6 2. 6. A number placed above another number, a little to the right, is called an exponent. Thus, in the expresSiOllS, 62, 73, 2 and 3 are exponents of 6 and 7, respectively. 7. An exponent placed over a quantity, denotes that the quantity is to be used as a factor as many times as there are units in the exponent. Thus, 24 = 2 x 2 x 2 x 2 =16. 8. When the exponent is a, the result is called the second power of the quantity over which it is placed. Thus, 72 = 7 x 7 =49=the second power of 7. 9. When the exponent is 3, the result is called the third power of the quantity over which it is placed. Thus, 4 3 = 4 x 4 x 4 = 64-_ the third power of 4. The higher powers are denoted in the same way. 10. The symbol V/, denotes that the square root of the quantity over which it is placed is to be taken. Thus, V/4=2; which is read, the square root of four equals two. 12 HIGHER ARITHMETIC. 11. The symbol V/, denotes, in a similar manner, the cube root of the number over which it is placed. Thus, V64=4; which is read, the cube root of sixty-four equals four. The roots of higher dimensions are denoted in a similar way. 12. The symbol.'. is equivalent to the phrase there fore or consequently. Thus 62 = 36, and 4 x 9 = 36.'. 6a=4 x 9; which is read, the square of six equals thirty-six, and the product of four and nine equals thirtysix; therefore, the scuare of six equals the product of four and nine. 13. The parenthesis, ( ), when it incloses several quantities, requires these quantities to be regarded as one single quantity. Thus, (5+3)x7-=56; which is read, the sum of five and three multiplied by seven equals fifty-six. EXAMPLES ILLUSTRATING THE FOREGOING.DEFINITIONS AND SYMBOLS. 3. THE expression, 11 + 5-2=- 2 x 7 = 28 - 2= 14, when translated into common language, becomes the sum of eleven and five diminished by two, equals the product of two and seven, equals twenty-eight divided by two, equals fourteen. 42 —8 2. The expression, 422 + 3 = 4 x 5 _ 20, is equivalent to the following:-forty-two diminished by eight, and the remainder divided by two, and the quotient increased by three, equals four multiplied into five, equals twenty. ILLUSTRATING DEFINITIONS AND SYMBOLS. 13 3. The expression, V/44 - 3 x 4 = 36 3 = 12, is the same as the square root of one hundred and forty-four, equals three multiplied into four, equals thirty-six divided by three, equals twelve. 4. Translate the expression, (10 +3)x7=182 +291, into common language. Ans. The sum of ten and three multiplied by seven is equal to one hundred and eighty-two divided by two, which is equal to ninety-one. 5. Translate the expression, (/1+ 7)x4= 4 V64 x 11, into common language. Ans. The square root of sixteen increased by seven, and the sum multiplied by four, is equal to the cube root of sixty-four multiplied by eleven. 6. Translate the expression, ( V/4 — -V) x 3 20 — 11, into common language. Ans. The square root of forty-nine diminished by the cube root of sixty-four, and the difference multiplied by three, is equal to twenty diminished by eleven. 7. What expression is equivalent to the following: "Five times nine divided by three, and that quotient multiplied by seven, equals the square of ten increased by five?" 5x9 Ans. 5 97=102 5. 8. What expression is equivalent to the following. "Three times twenty-one, increased by five times seven, and diminished by three times the square of four, is equal to twice the square of five?" Ans. 3x21+5x7-3x42=2x52. 9. What expression is equivalent to the following: "The cube root of sixty-four, increased by two, and 2 14 HGIIGER ARITInIETIC. the sum multiplied by ten, is equal to the square of eight diminished by four?" Ans. (/64-+ 2)x 10=82 -4. MULTIPLICATION OF COMPOUND EXPRESSIONS. 4. IN multiplication, the multiplier must always be regarded as an abstract number. The multiplicand may be a quantity or denominate number of any kind. Thus, we may repeat $7 once, twice, thrice, or any number of times, but we could not multiply dollars by yards, pounds, hours, or by any other denominate number. It is sometimes proposed to multiply money by money, as 2 s. 6 d. by 2 s. 6 d., but this is not philosophically correct. We may repeat 2 s. 6 d. once, twice, &c., but we cannot repeat it 2 s. 6 d. times. In estimating the cost of 20 bushels of apples at 25 cents per bushel, we do not repeat 25 cents 20 bushels of times, neither do we repeat 20 bushels 25 cents times; but, having fixed upon a quantity of apples equal to one bushel as our unit, we find that the quantity whose cost is to be estimated is 20 bushels, or twenty times the unit; now as one bushel is worth 25 cents, 20 bushels, being 20 times as many apples, must be worth 20 times as many cents as one bushel; we therefore repeat 25 cents twenty times, not 20 bushels of times, and thus obtain 500 cents, or 5 dollars, for the whole cost. And, in a similar way, we might show that, in all operations of multiplication, the multiplier is an abstract number, denoting how many times the multiplier is to be repeated. The product must evidently be of the same name as the MULTIPLICATION OF COMPOUND EXPRESSIONS. 15 multiplicand, since repeating a quantity once, twice, thrice, or any number of times, cannot change its denomination. jometirrncs the multiplicand is also an abstract number. It would be equally absurd to suppose the multiplier to be a negative quantity; for we could not repeat a quantity a minus number of times. If, then, in the course of an operation, we have for factors a positive and a negative quantity, we must regard the positive factor as the multiplier, and the negative quantity as the multiplicand; as, for example, if we wish the product of 4 and -7, we must repeat -7 four times, and the result will still be negative, we shall thus obtain - 28. Or we might have multiplied the 4 by 7, and changed the sign of the product. From which we see that when a minus quantity occurs in the multiplier, we are to multiply by it considered as positive, and then to change the sign of the product. Applying this principle to the case when both factors. are negative, as, for example, -5 multiplied by -7. In this case, it will be necessary to repeat — 5 seven times, and then to change the sign of the product, we thus obtain for our result, 35. From what has been said and done, we have for the multiplication of compound expressions the following RIULE. Multiply each term of one of the factors by each term of the other factor, observing that like signs produce plus, and unlike signs produce minus. Let it be required to multiply 3 + 2 by 4 + 5. We must repeat 3+2 as many times as there are units in 4+5. 16 lHIGI HER ARI'rTHnIETIC. First, repeating 3 + 2 as many times as there are units in 4, we get (3+ 2) x 4 = 12 + 8, for first partial product: Secondly, repeating 3+2 as many tones as there are units in 5, we get (3 + 2) x 5 = 15 + 10, for second partial product: Hence, 3+2, repeated as many times as there are units in 4+5, becomes (3+2) x(4+5)=12+8+15 + 10. Again, let it be required to multiply 7 —3 by 4+2: Proceeding as in the last example, we find (7 —3)x (4+2)=28- 12+ 14-6. In a similar way, we find that 4-3, multiplied by 3-2, gives (4-3)x(3-2)=12-9-8+6. EXAMPLES. 1. What is the product of 8+3 by 6 4? Ans. 48+18+32+ 12. 2. What is the product of 6 —2 by 4+3? Ans. 24-8+18-6. 3. What is the product of 11 —3 by 13-7? Ans. 143-39-77+21. 4. What is the product of 3+2 —1 by 4 —1+5? Ans. 12+8-4-3-2+1 - 15+10-5. 5. What is the product of 1+2-3 by 4-5+6? Ans. 4+8-12-5-10+15+6+12-18. 6. What is the product of 7-9 by 5-11? Ans. 35-45-77+99. 7. What is the product of 21-3 by 9-2? Ans. 189 —27 —42+6. 8. What is the product of 1 + 7 + 5 by 2 + 3? Ans. 2+14 + 10+3+ 21 ~+15. PRIME NUMBERS. 17 SINGULAR PROPERTY OF THE FIGURE 9. 5. Every number will divide by 9, when the sum of its digits is divisible by 9. For, take any number, as 78534; this number is, by the nature of decimal arithmetic, the same as 70000+ 8000+500+30+4. Now, 70000 = 9999 x 7+ 7 8000= 999x8+8 500= 99x5+5 30- 9x 3+3 4= +4.'. 78534=9999 x 7+999 x 8+99 x 5+9 x 3+(7+8+ 5+3+4.) Now, since each expression, 9999 x 7, 999 x 8, 99 x 5, and 9 x 3, is divisible by 9, it follows that the first number, 78534, will be divisible by 9 when the sum of its digits (7+8+5+3+4) is. Hence, it follows that any number being diminished by the sum of its digits, will become divisible by 9. Also, any number divided by 9, will leave the same remainder as the sum of its digits when divided by 9. The above properties belong to the digit 3, as well as to that of 9, since 3 is a divisor of 9. No other digit has such properties. NOTE.-These singular properties of the digit 9, have been made use of by many authors for proving the work of the four fundamental rules of arithmetic. PRIME NUMBERS. 6. No even number can, with the single exception of the number 2, be a prime, since all even numbers are divisible by 2. It is also evident that there are many odd numbers which are not primes. If we write in order 2* 18 HIGHER ARITHMETIC. the natural series of odd numbers, we discover that every third term, counting from 3, is divisible by 3; every fifth term, counting from 5, is divisible by 5; every seventh term, counting from 7, is divisible by 7, and so on. Commencing at 3, under every third term, I have placed a small figure, 3, to denote that the term under which it is placed is divisible by 3. Under every fifth term, counting from 5, I have, in like manner, placed a small 5, indicating that the corresponding term is divisible by 5. I have proceeded in the same way for the higher primes. Now it is evident that all the terms, under which there are no small figures found, are primes. We may also remark, that the numbers expressed by the small figures are the different prime factors of the numbers under which they are placed. 1, 3, 5, 7, 93, 11, 13, 153.5, 17, 19, 213.7, 23, 255, 273, 29, 31, 333.11, 355.7, 37, 393.13, 41, 43, 453.5, 47, 497, 51 3,1 53, 555.11, 573.19, 59, 61, 633., 655.13, 67, 693.3, 71, 73, 75.5, 777.11 79, 813, 83, 855 17 873.2,9 89, 917.,13 93331 95 5.19, 97, 993.,1 1 &c. In the above operation, we have found the primes only which are less than 100; but this process may be extended as far as we wish. This method of finding the successive primes was employed by Eratosthenes, who inscribed the series of odd numbers upon parchment, then cutting out such numbers as he found to be composite, his parchment with its holes resembled somewhat a sieve; hence, this method is called Eratosthenes' Sieve. The number 2, although an even number, must be regarded as coming under our definition of a prime, since the only number which will divide it is itself. PIRIME NUMBERS. 19 TABLE OF PRIME NUMBERS. 1!157137316071 8571110311399 163711949 2239 2531 2 163 379 613 859 1109 1409 1657 1951 2243 2539 3167 383 617 863 1117 1423 1663 1973 2251 2543 5 173 389 619 877 112311427 1667 1979 2267 2549 7 179 3971631 881 1129 1429 1669 19S7 2269 2551 11 181 401 641 883 1151 1433 1693411993 2273 2557 13119114091643 887 1153 1439 169711997 2281 2579 17 193 419 647 907 1163 1447 1699 1999 2287 2591 19 197 421 653 911 1171 1451 1709 2003 2293 2593 231199 431 659 919 1181 1453 1721 2011 2297 2609 29 2111433 661 929 1 87 1459 1723;2017 2309 2617 31 223 4391673 937 1193 1471 1733 202712311 2621 37 227 4431677 941 120 i 1481 1741 2029 2333 2633 41 229 449 683 947 121 3 1483 1747 203912339 2647 43 233 4571691 953 1217 1487 175312053 2341 2657 47 239 4611701 967 1223 1489 1759 2063 2347 2659 531241 463 709 971 1229 1493 l777 2069 2351 2663 59 251 467 719 977 1231 1499 17832081 2357 2671 61 257 479 727 983 1237 1511 1787 2083!2371 2677 67 263 4S87733 991 1249i 1523 1789 2087 2377 2683 7112691491 739 997 1259 1531 180112089 2381 2687 73/271 499 743 1009 1277 1543 1811 2099 2383 2689 79 2771503 751 1013 1279 1549 1823 211112389'2693 83 281 1509 757 1019 1283 1553 1831 2113 2393 2699 89 283 521 761 1021 1289 1559 1847 2129 2399 2707 971293 523 769 1031 1291 1567 1861 2131 2411 2711 101 307 541 773 1033 1297 1571 1867 21372417 2713 1031 311 547787 1039 11301 1579 1871 2141 2423 2719 107 3131557 797 1049 1303 1583 1873 2143i2437 2729 1093171563 809 1051 107 1597 1877 215312441 2731 113 331 569 811 1061 1319 1601 1879 2161 2447!2741 127 3371571 821 1063 1321 1607 1889 2179 2459 2749 131 347 577 82311069 1327 160911901 2203 2467 2753 137 349 587 827 10871361 1613 1907 2207 247312767 139t3531593 829 1091 1367 1619 1913 2213'2477 2777 149 359159918391093,1373 162111931 2221 2503'2789 151 367601 853 109711381 1627 193312237 2521'2791 20 HIGHIER ARITIHMETIC. The preceding table contains all the primes which are not greater than 2791. All prime numbers, except 2, are odd, and, therefore, terminate with an odd digit. Any number which ends with 5, is divisible by 5; hence it follows that all primes, except 2 and 5, must end with one of the figures, 1, 3, 7, or 9. When it is required to determine whether a given number is a prime, we first notice the terminating figure; if it is different from 1, 3, 7, or 9, the number is a composite; but if it terminate with one of the above digits, we must endeavor to divide it by some one of the primes, as found in the table, commencing with 3. There is no necessity of trying 2, for 2 will divide only the even numbers. If we proceed to try all the successive primes of the table until we reach a prime which is not less than the square root of the number, without finding a divisor, we may conclude with certainty that the number is a prime. The reason why we need not try any primes greater than the square root of the number, is drawn from the following consideration: If a composite number is resolved into two factors, one of which is less than the square root of the number, the other must be greater than the square root. The square of the last prime given in our table is 7789681; hence, this table is sufficiently extended to enable us to determine whether any number not exceeding 7789681 is a prime. It is obvious that numbers may be proposed which would require by this method very great labor to determine whether they are primes, still this is the only sure and general method as yet discovered. PRIME FACTORS. 21 Tables have been calculated, giving not only all the primes up to 3036000, but also the least prime factor of the composite numbers up to the same extent. Our table is of sufficient extent to enable us to work all ordinary examples. Any prime number, except 2 and 3, when divided by 6, must have a remainder of 1 or 5; for all prime numbers are odd, and any odd number when divided by an even number, must leave an odd number for remainder. Hence, any odd number divided by 6, must give 1, 3, or 5, for a remainder; if the remainder is 3, the number must have been divisible by 3, since the divisor and remainder are each divisible by 3. Hence, the remainder found by dividing a prime by 6, is 1 or 5. Therefore, by either adding one or subtracting one from any prime number greater than 3, it becomes divisible by 6. 7. Every number is either a prime number, or composed of prime factors. For, all numbers which are not prime are composite, and can, therefore, be separated into two or more factors; and, if these factors are not prime, they can again be separated into other factors, and thus the decomposition can be continued until all the factors are prime. Hence, to resolve any composite number into its prime factors, we have this R UL E. Divide the number by any prime number which will divide it without any remainder; then divide the quotient in the same way, and so continue until a quotient is 22 HIIGHER ARlTITIMETIC. obtained which is a prime. Then will the successive divisors, together with the last quotient, be the prime factors required. EXAMPL, ES. 1. Resolve 728 into its prime factors. Operaliom. 2 728 21364 21182 7 91 13 Therefore, 2x 2x 2x 7x 13= 23 x 7x 13, are the prime factors of 728. 2. Resolve 812 into its prime factors. Ans. 22x7x29. 3. What are the prime factors of 978? Ans. 2x3x 163. 4. What are the prime factors of 1011? Ans. 3 x 337. 5. What are the prime factors of 100? Ans. 22 x 52. 6. What are the prime factors of 8975? Ans. 52 x 359. 7. What are the prime factors of 808? Ans. 23X 101. 8. W"hat are the prime factors of 707? Ans. 7x 101. 9. What are the prime factors of 1118? Ans. 2x 13x 43. PRl'lIME FA'ACUTORS. 23 S. Suppose we wish to know whether the numbers 204 and 468 have a common factor, we proceed as follows: We decompose them into their prime factors, and thus obtain 204=22 x 3 x 17, and 468= —22 x 32 x 13. Here we see that 22 x 3 is common to both the numbers 204 and 468. Hence, to find the greatest factor which is common to two or more numbers, or, as generally expressed, to find the greatest common measure of two or more numbers, we have this RULE. Resolve the numbers into their prime factors, (by Rule under Art. 7.) Then select such of the primes as are common to all the numbers, multiply them together, and the product will be the greatest common measure. EXAMPLES. 1. What is the greatest common measure of 1326, 3094, and 4420? These numbers, when resolved into their prime factors, become 1326=2 x3x13 x 17 3094=2 x7x13x17 4420=22 x 5 x 13 x 17 The factors which are common, are 2, 13, and 17; therefore, the greatest common measure is 2 x 13 x 17= 442. 2. What is the greatest common measure of 556, 672, and 840? Ans. 22=4. 3. What is the greatest common measure of 110, 140, and 680? Ans. 2x5= 10. 4. What is the greatest common measure of 255, and 532? Ans. They have none. 24 HIGHER ARITHMETIC. 5. What is the greatest common measure of 375, 408, and 922? Ans. They have none. 9. We may also find the greatest common measure of two numbers by the following RULE. Divide the greater by the less, then divide the divisor by the remainder, and thus continue to divide the preceding divisor by the last remainder, until there is no remainder. The last divisor will be the greatest commQzn measure. NoTE. —Where there is no common measure, the last divisor will be 1. EXAMPLES. 1. What is the greatest common measure of 360 and 630? Operation. 360)630(1 360 270)360(1 270 90)270(3 270 Hence, the greatest common measure is 90. 2. What is the greatest common measure of 922, and 408? Ans. 2. 3. What is the greatest common measure of 1825, and 2555? Ans. 365. LEAST COMMON MULTIPLE. 25 4. What is the greatest common measure of 124, and 682? Ans. 62. 5. What is the greatest common measure of 296, and 407? Ans. 37. 6. What is the greatest common measure of 404, and 364? Ans. 4. 7. What is the greatest common measure of 506, and 308? Ans. 22. 8. What is the greatest common measure of 212, and 416? Ans. 4. 9. What is the greatest common measure of 74, and 84? Ans. 2. 10. Suppose we wish to know what is the least number which will divide by 215 and 460; we proceed as follows: We decompose them into their prime factors, and thus obtain 215-5 x 43, 460=22 x 5 x 23. Hence, we see that 22 x 5 x 23 x 43 =- 19780, is the least number which can be divided by 215 and 460. Hence, to find the least number which will divide by two or more numbers, or, as generally expressed, to find the least common multiple, we have this RULE. Resolve the numbers into their prime factors, (by Rule under Art. 7); select all the different factors which occur, observing when the same factor has different powers, to take the highest power. The continued product of the factors thus selected will be the least common multiple. EXAMPLES. 1. What is the least common multiple of 12, 16, and 24? HIGHER ARITHMETIC. These numnjers, resolved into their prime factors, give 12-22 x 3 16=24 24= —23 x 3 Therefore, 24 x 3=48 is the least multiple required. 2. What is the least common multiple of 9, 12, 16, 20, and 35? Ans. 24 x 32 X 5 x 7=5040. 3. What is the least common multiple of 7, 13, 39, and 84? Ans. 22x3x7x 13=1092. 4. What is the least common multiple of the nine digits? Ans. 23 X 32 X 5x 7-2520. 5. What is the least common multiple of 3, 5, 7, 12, 15, 18, and 35? Ans. 22 X 32 X 5 x 7= 1260. 6. What is the least common multiple of 100, 109, 463, and 900? Ans. 22 x 32 x 52 x 109 x 463 =45420300. 7. What is the least common multiple of 365, 910, 2217, and 2424? s 23 x3x 5x7x 13x 73x 101 x739As. 59499225240. 11. We may also find the least common multiple of two or more numbers by the following RULE.* Write the numbers in a horizontal line; divide them by any prime number which will divide two or more of them; place the quotients with the undivided terms for * This rule is usually given as follows: "W rite down the numbers in a line, and divide them by any number that will measure two or more o:' them, and write the quotients and undivided numbers in a line beneath. Divide this line as before, and so on, until LEAST COMMON MULTIPLE. 27 a second horizontal line; proceed with this second line as with the first, and so continue, until there are no two terms which can be divided. The continued product of the divisors and the numbers in the last horizontal line will be the least common multiple. EXAMPLES. 1. What is the least common multiple of 28, 35, 42, 77, and 70? Operation. 7 28, 35, 42, 77, 70 5 4, 5, 6, 11, 10 2 4, 1, 6, 11, 2 2, 1, 3, 11, 1 Hence, 7x5x2x2x3x11=4620, is the multiple sought. there are no two numbers that can be measured by the same divisor; then the continual product of all the divisors and numbers in the last line will be the least common multiple required:" The above we have copied from Mr. Adams' Arithmetic. Nearly all our Arithmetics give, in substance, the same rule. We will now show, by an example, that this rule may give very different results, depending upon the divisors used, and of course the rule is in fault. EXAMPLE. What is the least common multiple of 12, 16, and 24 We will work this example in three ways, as follows: First Operation. Second Operation. Third Operation. 12 12, 16, 24 8 12, 16 24 4 12, 16, 24 2 1, 16, 2 3 12, 2, 3 3 3, 4, 6 1, 8, 1 2 4, 2, 1, 2 1, 4, 2 I 2, 1, 1 1, 2, 1 12x2x8=192. 8x3 x2x2=96. 4x3x2x2=248. These operations, which are wrought strictly by this rule, give 192, 96, and 48, for the least multiple of 12, 16, and 24. Hence, the rule is wrong, and cannot be depended upon. The least common multiple of 12, 16, and 24, is 48, as may be found by either of our rules. 28 HIGHER ARITHMETIC. In the preceding example, we find, by inspection, that all the given numbers, 28, 35, 42, 77, 70, are divisible by 7, giving, for the second horizontal line, the numbers 4, 5, 6, 11, 10. Now 7 times the least multiple of 4, 5, 6, 11, 10, is the least multiple of 28, 35, 42, 77, 70, since the latter numbers are respectively 7 times the former. Again, of the numbers 4, 5, 6, 11, 10, we find that 5 and 10 are divisible by 5. Dividing, we find for the third horizontal line, the numbers 4, 1, 6, 11, 2; now, as before, 5 times the least multiple of 4, 1, 6, 11, and 2, is the least multiple of the numbers of the second line. Again, of the numbers 4, 1, 6, 11, 2, we find that 4, 6, and 2, are divisible by 2. Dividing, we obtain for the fourth horizontal line, 2, 1, 3, 11, 1; and, as before, twice the least multiple of the last numbers is the least multiple of 4, 1, 6, 11, 2, which, multiplied by 5, gives the least multiple of 4, 5, 6, 11, 10; and this result being multiplied by 7, gives the least multiple of the numbers sought. When the division is continued until there are no two terms which can be divided, the continued product of the numbers constituting the last horizontal line is the least multiple. Hence the correctness of the rule. 2. What is the least common multiple of 46, 92, 374, and 23? Ans. 22 x 23 x 187=17204. 3. What is the least common multiple of 5, 15, 36 and 72? Ans. 23 x 32 x 5-360. 4. What is the least common multiple of 11, 77, 88, and 92? Ans. 23 x7x 11 x 23=14168. 5. What is the least common multiple of 14, 51, 102, and 500? Ans. 2 x 3 x 7 x 17 x 250=- 178500. T'1'E DIVISORS oF NUMBERS. 29 12. Suppose we wish to find all the divisors of 36, we proceed as follows: We resolve 36 into its prime factors, and thus obtain 36 —22 x 32. Now it is obvious that any combination of 2 and 3, which does not make use of these factors in a higher power than they occur in 22 x 32, must be a divisor of 36. All such combinations can be found by multiplying 1 +2+4 by 1+3+9. Performing this multiplication, we obtain 1 +2+4 1+3+9 1 +2+4+3+ 6+ 12+9+ 18+36. Therefore, the divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36. Hence, to find all the divisors of any number, we have this RULE. Resolve the number into its prime factors; form as many series of terms as there are prime factors, by making 1 the first term of any one of the series, thefirst power of one of the prime factors for the second term, the second power of this factor for the third term, and so on, until we reach a power as high as occurred in the decomposition. Then multiply these series together, (by Rule under Art. 4,) and the partial products thus obtained will be the divisor sought. EXAM PL ES. 1. What are the divisors of 48? Here we find 48=24 x 3. Therefore, our series of terms will be 1 + 2 + 4 + 8 + 16 and I + 3; multiplying these together, (by rule under Art. 4,) we get 3' 30 IIGIHER ARITIIMETIC. 1 +2+4+8+ 16 1+3 1+2+4+8+ 16+3+6+12+24+48. Therefore, the divisors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48. 2. What are the divisors of 360? Ans. ( 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24,. 30, 36, 40, 45, 60, 72, 90, 120, 180, 360. 3. What are the divisors of 100? Ans. 1, 2, 4, 5,10, 20, 25, 50, 100. 4. What are the divisors of 810? Ans. i1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 81, 90, 135, 162, 270, 405, 810. 5. What are the divisors of 920? Ans. 1,2, 4, 5, 8, 10, 20, 23, 40, 46, 92, 115, 184,. 230, 460, 920. 6. What are the divisors of 840? v 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, Ans.! 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840. 7. What are the divisors of 1000? Anis. 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200,, 2 50, 500, 1000. NOTE.-We may observe that all the divisors of these respective divisors are included in the same series, since they must evidently be divisors of the original number. 13. Since the series of terms which we multiplied together by the last rule, to obtain the divisors of any number commenced with 1, it follows that the numnrber of terms in each series will be one more than the units in the exponent of the factor used. THE DIVISORS OF NUMBERS. 31 Hence, to find the number of divisors of any number without exhibiting them, we have this R IL E. Resolve the number into its prime factors; increase each exponent by a unit, and then take their continued product, and it will express the number of divisors. EIXAMPLES. 1. How many divisors has 4320? 4320=25 x33 x 5. In this case, the exponents are 5, 3, and 1, each of which being increased by one, we obtain 6, 4, and 2, the continued product of which is 6 x 4 x 2=48, the number of divisors sought. 2. How many divisors has 300? Ans. 18. 3. How many divisors has 3500? Ans. 24. 4. How many divisors has 162000? Ans. 100. 5. How many divisors has 824? Ans. 8. 6. How many divisors has 1172? Ans. 6. 7. How many divisors has 6336? Ans. 42. 8. How many divisors has 75600? Ans: 120. 32 HIGHER ARITHIMETIC, CHAPTER II. FRACTIONS. 14. A FRACTION is an expression representing a part of a unit. VULGAR FRACTIONS. 15. A VULGAR FRACTION consists of two numbers, the one placed above the other, as in division. The number above the line is called the numerator; the number below the line is called the denominator. Thus, 5 is a vulgar fraction, whose numerator is 5, and whose denominator is 8; it is read five-eighths. The denominator shows how many parts the unit is divided into, and the numerator shows how many of these parts are used. Thus, A denotes that the unit is divided into 8 equal parts, and that 5 of these parts are used. When the numerator is equal to the denominator, the fraction is equivalent to a unit. Thus,,,6 1, 5 and 3 1 are each equivalent to 1. When the numerator is less than the denominator, the value of the fraction is less than a unit; it is then called a proper fraction. Thus, 5-, 7, I, and 2, are'each proper fractions. When the numerator is larger than the denominator, its value is then more than a unit; it is, therefore, called an improper fraction. Thus, 8, 7, -1-I, and 4, are each improper fractions. RE DUCTION OF FRACTIONS. 33 A fraction of a fraction, connected by the word of, is called a compound fraction. Thus, 2 of 5 of -2 of,7 2 of 2 of A, and -2 of 1 of 3a, are compound fractions. A fraction is said to be inverted, when the numerator and the denominator change places. Thus, the fractions I7, 1 1 and 5, when inverted, become 1, 7 -, and 8. Any integer may take the form of an improper fraction, by writing a unit for its denominator. Thus, 6, 5, 3, and 11, are the same as the improper fractions 6 5 3 and -'-. A number consisting of an integer and a fraction, is called a mixed number. Thus, 41, 5 7, 6 3, and 131 7, are mixed numbers. They may also be written 4+-, 5 + 7, 6 +, and 13+ 1. REDUCTION OF FRACTIONS. 16. SINCE the value of a fraction is the quotient arising from dividing the numerator by its denominator, we may infer the following Propositions. I. That, multiplying the numerator of a fraction by any number, is the same as multiplying the value of the fraction by the same number. II. That, multiplying the denominator of a fraction by any number, is the same as dividing the value of the fraction by the same number. 34 1;'' III. That, multiplying both numlerator and denominator by the same number, does not alter the value of the fraction. IV. That, dividing the numerator of a fraction by any number, is the same as dividing the value of the fraction by the same number. V. That, dividing the denominator of a fraction by any number, is the same as multiplying the value of the fraction by the same number. VI. That, dividing both numerator and denominator by the same number, does not alter the value of the fraction. l7. When the numerator and denominator of a fraction have no common measure, it is said to be in its lowest terms. To reduce simple fractions to their lowest terms,* we have the following *The following properties may frequently be of assistance in abbreviating vulgar fractions: 1. If any number terminate on the right with zero, or an even digit, the whole will be divisible by 2. 2. If any number terminate on the right with zero, or 5, the whole will be divisible by 5. 3. When the number expressed by the two right-hand figures are divisible by 4, the whole will be divisible by 4. 4. When the number expressed by the three right-hand figures are divisible by 8, the whole will be divisible by 8. 5. If the sum of the digits of any number be divisible by 3 or 9, then the whole number will be divisible by 3 or 9. This has already been shown under Art. 5. 6. When the difference between the sum of the digits occupying the odd places, counting from the right towards the left, and the sum of the digits occupying the even places is zero, or divisible by 1I, then the number will be divisible by 11. REDUCTION OF FRACTIONS. 35 RULE. Divide both numerator and denominator by their greatest common measure (found by one of the Rules under Art. S, or 9.) This division will not alter the value of the fraction. (Prop. VI., Art. 16.) EXAMPLES. 1. Reduce 43 7 5 to its lowest terms. In this example, we find the greatest common measure of 375 and 425 to be 25. Dividing both numerator and denominator by 25, we find 432 = 1. 2. Reduce 04 9 to its lowest terms. Ans. 8. 3. Reduce -— 728 to its lowest terms. Ans. A. 4. Reduce 740 o to its lowest terms. Ans. -. 5, Reduce 7 2 5 4 6 to its lowest terms. Ans..3 7 9 8 6. Reduce 9 0 62 to its lowest terms. Ans. 4 5 0 7. Reduce -4 to its lowest terms. Ans. a-l'. 8. Reduce -4-Y-8' to its lowest terms. Ans. -23-. 9. Reduce 4345. to its lowest terms. Ans. 691 Were we to resolve the numerator and denominator into their prime factors, we should then at once discover the factors common to the numerator and denominator, and could, therefore, strike them out, and then the fraction would be in its lowest terns. For example, let it be required to reduce the fraction ~72. Resolving the numerator and denominator into their prime factors, the fraction will become 7 2 2 3x 7 x 13 2_ -x1- 3. Here we discover that the factors 22 x 7 22 x7x29 are common to both numerator and denominator. Striking them out, we obtain 36 HIGHER AR'IIMETI C, 728 23x7x13 2x13 26 812 22x7x29 29 -29 In a similar way, we find 210 2x3x5x7 2><7 14 495 32x5x 11 3xl1 33 735 3x5x72 3 2695 5 x 72 x 11 11 It is obvious that this method may be used in all cases. Whenever we discover, by inspection, any number which will divide both numerator and denominator without a remainder, we may use it as a divisor, before resorting to either of the above methods. 18. To reduce an improper fraction to a mixed number, we have this RUL E. Divide the numerator by the denominator; the quotient will be the integral part of the mixed number. The remainder, placed over the denominator of the improper fraction, will form the fractional part. The correctness of the above rule is obvious from considering that the value of a fraction is the quotient arising from dividing the numerator by the denominator. EXAMPLES. 1. Reduce J4-1 to a mixed number. Dividing 17 by 4, we obtain the quotient 4, with the remainder 1;.. the mixed number equivalent to gJ-7, is 4~, or 4 +. REDUCTION OF MIXED NUMBERS. 37 2. What mixed number is equivalent to 1 - 1? Ans. 18g. 3. What mixed number is equivalent to J-2I-3-45? AyAs. 1122-1-'. 4. What mixed number is equivalent to 2 9 "7? Ans. 812 -2 5. What mixed number is equivalent to -76%%-9%3? Ans. 11 635 6. What mixed number is equivalent to -327-9? Ans. 1123 5 7. What mixed number is equivalent to -37.9? Ans. 6412A. 8. What mixed number is equivalent to 4-8-59793-5-4? Ans. 1294~237 - 328=1294-2 869 2 9. What mixed number is equivalent to 223_39? Ans. 52433. 10. What mixed number is equivalent to ~2_i07? Ans. 4955. 11. What mixed number is equivalent to *3659773? Ans. 12 049 8 1 19. To reduce a mixed number to its equivalent improper fraction, we have this RULE. Mu10.ltiply the integral part of the mied number byis equivalent to the denominator of the fractional part; to the product add the numerator of the fractional part; the sum will be the numerator of the improper fraction, under which place the denominator of the fractional part. This rule is obviously correct, since it is the reverse 4 38 IIIGHER ARITHIMETIC. of the rule under Art. 18, where a reverse operation was required to be performed. EXAMPLES. 1. Reduce 13A to an improper fraction. Multiplying the integer 13 by the denominator 7, we obtain 91; to which, adding the numerator 6, we get 97 for the numerator of the improper fraction;.'. the improper fraction equivalent to 137 is —. 2. What improper fraction is equivalent to 1278-? Ans. 3 83 55 3. What improper fraction is equivalent to 189104? Ans. 1 32 7 4. 4. What improper fraction is equivalent to 492536~1? Ans. 6402979 5. What improper fraction is equivalent to 5878? Ans. -96644 A ns. 387883 6. What improper fraction is equivalent to 11 -42-3? Ans. 3-5698 7. What improper fraction is equiva lclnt to 23,23? Ans. -53594 2329 8. What improper fraction is equivalent to 375- I? Ans. -89'1. 9. What improper fraction is equivalent to the mixed number 68334-7-U?-? Ans. 30505o3072 10. What improper fraction is equivalent to the mixed number 11223344-5? Ans. 1 2 3 456 7 8 9 20. Reduce the compound fraction 3 of I7 to its equivalent simple fraction. 4 of -7T can be obtained by dividing the value of the REDUCTION OF COMPOUND FRACTIONS. 39 fraction -7 by 4, which (by Prop. II., Art. 16,) can be effected by multiplying the denominator by 4; 1 7 7 o4 11 4x11 Again, 4 of,-7T is obviously three times as great as 7 ~ of 1;'. to obtain 3 of -T, we must multiply 4 by 3, which (by Prop. I., Art. 16,) can be done by multiplying the numerator by 3; hence, we have 3 of 7 3x7 21 11 4x 11 44 Hence, to reduce compound fractions to their equivalent simple ones, we have this RULE. Consider the word OF, which connects the fractional parts as equivalent to the sign of multiplication. Then multiply all the numerators together for a new numerator, and all the denominators together for a neiw denominator, always observing to reject or cancel such factors as are common to the numerators and denominators, which is the same as dividing both numerator and denominator by the same quantity, and (by Rule under Art. 17,) does not change the value of the fraction. EXAMPLES. 1. Reduce - of 4 of ~, of - to its equivalent simple fraction. Substituting the sign of multiplication for the word of, we get I x 3x i8 X -2. First canceling the 8 of the 40 HIGHER ARITHMETIC. numerator against the 2 and 4 of the denominator, by 1 3 5 drawing a line across them, we get x x - x 12 Again, canceling the 3 and 5 of the numerator against the 15 of the denominator, we finally obtain 1 $ $ $ 1 xx4 x 12=12 2. Reduce 3 of l-o4 of I of 4 of T-5 to its simplest form. First, canceling the 7 and 5 of the numerator against 3 14 2 4 $ the 35 of the denominator, we get X x x 1 Again, canceling the 7 of the denominator against a part of the 14 of the numerator, and the 3 of the numerator against a part of the 9 of the denominator, we obtain 2 8 4 4 8$ $ 8 a 11 3 Finally, canceling the 2 and 4 of the numerator against 8 of the denominator, we get 8 4 4 $ 1 ~x -x-x x 7 $$ s 9$ 11 33 3 NOTE. —We have written our fractions several times, in order the more clearly to exhibit the process of canceling. But in practice, it will not be necessary to write the fraction more than once. It will make no difference which of the factors are first canceled. When all the common factors have, in this way, been stricken out, the fraction will then appear in its lowest terms. The student will find it to his interest to perform many examples of this kind, as this principle of canceling will be extensively employed in the succeeding parts of this work. COMMON DENOMINATOR. 41 3. Reduce -I3r of - of 24 6 of 43 to its simplest form. Ans. V-f8 4. Reduce I of 7 of 1 of _24 of 4 of 4T of 5 to its simplest form. Ans. -1 5. 5. Reduce A3-q of 29 of 1-8 of 4 of to its simplest form. Ans. 2 6. Reduce I Iof 73 of 4 of'8 to its simplest form. Ans. 6292 7. Reduce 4 of 2 of - of 5 of to its simplest form. Ans. 536. 8. Reduce I of 2 of 3 of 4 of 5 of 4 of 7 of 4 to its simplest form. Ans.. 9. Reduce -27 of -'1- of 2 of 4 to its simplest form. Ans. TW9 1 - 10. Reduce 7 of 44 of I 4 of 34 to its simplest form. Ans. 1536. 21. To reduce fractions to a common denominator, we have this RULE. Reduce mixed numbers to improper fractions-compound fractions to their simplest form. Then multiply each numerator by all the denominators, except its own, for a new numerator, and all the denominators together for a comnmon denominator. It is obvious that this process will give the same denominator to each fraction, viz: the product of all the denominators. It is also obvious that the values of the fractions will not be changed, since both numerator and denominator 4* 42 i1IGIhER ARITHMETIC. are multiplied by the same quantity, viz: the product of all the denominators except its own. E X A M P I, 1: S. 1. Reduce I, of 2, l3, and,7 of 2, to equivalent fractions having a common denominator. These fractions, when reduced to their simplest form, are-, 2, 1 2 and 2. The new numerator of the first fraction is 1 x 3 x 11 x 9 = 297. The new numerator of the second fraction is 2 x 2 x 11 x 9-396. The new numerator of the third fraction is 3 x 2 x 3 x 9-162. The new numerator of the fourth fraction is 2 x 2 x 3 x 11-122. The common denominator is 2 x 3 x 11 x 9 -594. Therefore, the fractions, when reduced to a common denominator, are 2 9, 46, 24, and 3. 2. Reduce 4 of, 7-6 of -17-, and 44, to equivalent x,?1 7 4Jr fractions having a common denominator. Ans. 2 09 9 6 8, and 424.0'Rde, ~2 9 61d 3. Reduce - 3 of 37, and 44, to equivalent frac tions having a common denominator. Ans. 5-47, 6449 and 76467 4. Reduce 231 4, and 47 to equivalent fractions having a common denominator. Ans. 4 6 3 5 9 9 2 0 4 8 82 1 9 1799 2,. 2 5. Reduce 73, 8, and -7T, to fractions having a coinmon denominator. Ans. -4 -6 4'r, 7 "2, 8 -~ 6.7 7 1 Ru I, an {~0, 6. Reduce 7-4 7, and II 4a4, to fractions having a common denominator. Ans. 942779 1 9 3 T-j-97TT-D YW0D-T-FTT REDUCTION OF COMPOUND FRACTIONS. 43 22. To reduce fractions to their least common denominators, we have this It U L E. Reduce the fractions to their simplest form. Then find the least common multiple of their denominators, (by Rule under Art. 10, or Rule under Art. 11,) which will be their least common denominator. Divide this common denominator by the respective denominators of the given fractions; multiply the quotients by their respective numerators, and the products will be the new numerator. The correctness of the above rule may be shown in the same way as was that of the preceding rule. EXAMPLES. 1. Reduce 2- of 7 of -7,, and I-, to equivalent fractions having the least common denominator. These fractions, when reduced to their simplest form, become I, 23,, and -75. The least common multiple of the denominators 8, 20, and 15, is 120=common denominator. New numerator of the first fraction is -1 x 1 = 15. ic" " of the second fraction is 2-~ x 3=18. sc" " of the third fraction is -JT~-k x7=56. Hence, the fractions, when reduced to their least common denominator, become -2, JL8, and 52 2. Reduce 8 of 4, 4-, and -2, to equivalent fractions having the least common denominator. Ans. 14 9 o, and -3 3. Reduce 7 of 3 of 6 of 2 64 2, and 7-, to fractions having the least common denominator. Ans. 22,, and 250~ 2- 0'-... 44 HIGHER ARITHMETIC. 4. Reduce 1 of 439 of -4 of J12, 6, and 28, to fractions having the least common denominator. Ans. 2, 4-3, and 2 5. Reduce'& of 22 of - of 5 5 of and 3 of 2 8 2 2 4 9and,6 off 44, to fractions having the least common denominator. Ans. 1 7 6 1 2 0 1 9 4 4.81 422576 ans. {78 s368, 7606-368, 7 -o — 3 86. Reduce -v-, 11-7, and 7-53 to fractions having the least common denominator. Ans. -363 2706 1 652 7. Reduce _ of 4 of of40 of 3 3-, and 10-, to fractions having the least common denominator. Ans. 8 72 5 63 8. Reduce 3,4 3 of 4-5, and 3 7 to fractions having the least common denominator. Ans. - 99 2444 7030 9. Reduce 34, 54, 54, and 11 - to fractions having the least common denominator. Ans. 4840 8008 8085 173Te s. 40, 1oo4 0 1o540' 1\?ADDITION OF FRACTIONS. 23. SUPPOSE we wish to add 3 and 4. We know that so long as these fractions are of different denominators they cannot be added; we will, therefore, reduce them to a common denominator; we thus obtain 3= = 3, 4_ 23 Now, taking their sum, we get + - 5 + 2 15+28 4 31 8 35 — 3 5 35 Hence, to add fractions, we have this RULE. Reduce the fractions to a common denominator, and take the sum of the numerators, under which place the common denominator, and it will give the sum required. ADDITION OF FRACTIONS. 45 EX A M P,ES. 1. Add the fractions I of 5, P3 of -y-, and 54. These fractions, reduced to their least common denominator, are 5 7, and 7-; and their sum is 5+7+72 84 14 14 2. Add the fractions, 4, 7, and -., Ans. 6 8 = 2_4. 3. Add the fractions 45 of 7,,3, and 44. Ans. = 5A-54. Add the fractions 43,, 4 3 and i-. Ans. 44A=2.1857 3 7 4 0 3 7 4 7 5. What is the sum of 73 3 and 43? Ans. -9 1 6. What is the sum of 234,, 22 and 4? Ans. 1 4 3 5 3 7 26- 7 5 28 9. 7. What is the sum. of 449 243 and 7? Ans. 9 =3 =29 6 8. What is the sum of I of 44, - of 64, and 5? Ans. i —-= 3 3 7 9. What is the sum of, 4, - 6 and -1? Ans. -6-~ —o= 12. 10. What is the sum of, 2, 43 5, and,9? Ans. 4. 11. What is the sum of I,1 2,, 3-, 44, and 54? Ans. 16 -T. 46 HIG.HER ARITHMETIC. SUBTRACTION OF FRACTIONS. 24. To subtract one fraction from another, we have this RULE. Reduce the fractions to a common denominator, and subtract the numerator of the subtrahend from that of the miinuend; place the common denominator under the difference. EXAMPLES. 1. From -3,- subtract 331. These fractions, when reduced to their least common denominator, become 3 341. Therefore, T3TT81 93 88- 5 341 341 3412. From I73 subtract -2 Ans. 1?241. 3. From 27 subtract 1 9 Ans. 2 = 1 1 1 4. From 3 of 2- of 6 subtract I of,-. Ans. -72 5. From -37 subtract 4-3. Ans. I.2 6. From 49 subtract T7W3 Ans. T447784 T~3 0 63' I 64'TW 7. From I- subtract -4 Ans. 257 8. From 3 3 subtract 2. Ans. 11 9. From I of 41 subtract 3. Ans. 50 7. 10. From 22 of 429 subtract 3 of 2. Ans. 1 w79 3 11. From 3 of 2 subtract 29. Ans. -9~7 12. From 64 subtract 22 Ans. 44 4. MULTIPLICATION OF FRACTIONS. 47 MULTIPLICATION OF FRACTIONS. 5. LET it be required to multiply 4 by -. We have seen (under Art. 20,) that 3 multiplied by - is the same as 3 of 5. Therefore, we must use the same rule for multiplying fractions as for reducing compound fractions. Hence, to multiply together fractions, we have this RULE. Multily all the numerators together for a new numerator, and all the denominators together for a new denominator, always observing to reject or cancel such factors as are common to both numerators and denominators. EXAMP L ES. 1. Multiply together the fractions - T,, and 3. Expressing the multiplication, we obtain 1-3 x 22x x. Canceling the 3 and 7 of the numerators, against 21 of the denominators, also the 11 of the denominators against a part of the 22 of the numerators, we get 2'$ /I 1 2 2 X X- Xji^t 2 9 3 9X3 27 2. Multiply together the fractions -,, -, and 4. Indicating the multiplication, we get,-x x2 x A4i x 4. Canceling the 11 of the denominators, against a part of the 55 of the numerators, also the 7 of the numerators, against a part of the 35 of the denominators, we obtain 5 /- 21 I 4 X Vs X X 9x5 48 HIGHER ARITHMETIC. Again, canceling the 5, which is common to both numerators and denominators, also the factor 7, which is common to 21 of the numerators, and to 42 of the denominators, we get 3 $ ~ At $$ 4 X X Xl 6 Finally, canceling the 3 of the numerators, against a part of the 9 of the denominators, and the factor 2, which is common to the 4 of the numerators, and to the 6 of the denominators, we obtain $ 2 00 Ad 4 - 2 2 -X -x- x9-3x3 93 NOTE.-A little practice will enable the student to perform these operations of canceling with great ease and rapidity. And since, as was remarked under Art. 20, it is immaterial which factors are first canceled, the simplicity of the work must depend much upon his skill or ingenuity. 3. Multiply together the fractions 3,, and 4~ Ans. 48' ip,=,4 26 4. Multiply together the fractions 3, 74, 2, 11. Ans. -. 5. Multiply together the fractions, 3, 1- of 33, and Ans.- 2~ 0. 6. Multiply together the fractions 2, 7, and 4L Ans. 38. 7. Multiply together 7, 4', and 6. Ans. -. ~, 2~ 3 5 8. Multiply together -, -, 22, and ~. Ans. 9. 9. Multiply together 7, 8 1 3, nd d Ans.. DIVISION OF FRACTIONS. 49 10. Multiply together 2, X4,,, and 5. Ans. TT. DIVISION OF FRACTIONS. 26. LET it be required to divide 4 by 58. We know that 4 can be divided by 5, by multiplying the denominator by 5, (see Prop. II., Art. 16,) which 4 gives 4. Now, since 5 is but one-eighth of 5, it follows that 4, divided by 5, must be eight times as great as A4 divided 4x8 by 5.. 4, divided by 5, must be 7x5'From this, we see that 4 has been multiplied by -, when inverted. Hence, to divide one fraction by another, we have this RULE. Reduce the fractions to their simplest form. Invert the divisor, and then proceed as in multiplication. EXAMPLES. 1. Divide 4 3 by 28. Inverting the divisor, and then multiplying, we obtain 4 3 X 2 1; which, by canceling, becomes ~x 8 X4 8; 8 4 2 2. Divide 2-0 by 478v Ans. 4 —=30I. 3. Divide 34 by 9- Ans. 48-2 o-2-=2394. Divide 44 by 3. Ans. 3. 5. Divide 4} by 1712. Ans. 125. 50 HIGHER ARITHMETIC. 6. Divide 2 of 44 by 2 of 5. Ans. -1-3-5-4 4. 7. Divide 3 of 3 of 7 by J3 of 6. Ans. -8 8. Divide -4 by 92 Ans. 74. 9. Divide _ of 23 of -4 by 3 of 4. Ans..9 10. Divide ~- of -2'2 of - by g-. Ans. 17. COMPLEX FRACTIONS. 27. SOMETIMES fractions occur, in which the numerator, or denominator, or both, are already fractional. 2 2 3 1 Thus, - -X 4-,' -: such fractions are called complex fractions. REDUCTION OF COMPLEX FRACTIONS. 28. SINCE the value of a fraction is the quotient arising from dividing the numerator by the denominator, it follows that the complex fraction 2 is the same as 2 - = —? —=42. Again, =4-.- 7 6=. Hence, to reduce a complex fraction to a simple one, we have this RULE. Divide the numerator of the complex fraction by the denominator, according to Rule under Art. 26. EXAMPLES. 1. Reduce 4-} 1. Reduce to a simple fraction 33 Dividing 4}=: by 3>- -, we get 2 7 -. GIVEN DENOMINATOR. 51 2. Reduce -7 to a simple fraction. Ans. 9 3' 6 3. Reduce 37 to a simple fraction. Ans. ~ —=3+ 4. Reduce T to a simple fraction. 34 Ans. 1 — D-a 18 13 5. Reduce -4-1 to a simple fraction. Ans. 13 41 6. Reduce -6, to a simple fraction. Ans. 89. 8 8 7. Reduce to a simple fraction. of I~ Ans. _343 =84. 8. Reduce 24 -- to a simple fraction. Ans. -3 —=438 - of - - 6~' 9. Reduce 6+ to a simple fraction. Ans. Ao. 10. Reduce 10'x to a simple fraction. Ans. 7 3 9 a 3 2 REDUCTION OF FRACTIONS TO A GIVEN DENOMINATOR. 29. SUPPOSE we wish to change the fraction 4 to an equivalent one, having 6 for its denominator. It is obvious that if we first multiply 4 by 6, and then divide the product by 6, its value will not be altered. Bn5 6 _ By this means, we find that - x 6- _4 5 6 6 6 52 HIGHER ARITHMETIC. Hence, to reduce a fraction to an equivalent one having a given denominator, we have this RULE. Multiply the fraction by the number which is to be the given denominator, (see Rule under Art. 25,) under which place the given denominator, and it will be the fraction required. EXAMPLES. 1. Reduce 7 to an equivalent fraction having 8 for its denominator. In this example, we first multiply 7 by 8, which gives 24; therefore, placing 8 under 24, we get - for the fraction required. 2. Reduce l3 to an equivalent fraction having 12 for its denominator. 3T-3 Ans. - 12 3. Reduce 1 3 to an equivalent fraction having 7 for its denominator. Ans. Ans.'. 7 4. Reduce,,10 and to fractions having 12 for 2104 10 their common denominators. Ans. 12, -2, I,3, and -,-' T1 5. Reduce a, T, A, and Il, to fractions having 100 for their common denominator. Ans.' and 100 100' 100' 100j 100a REDUCTION OF DENOMINATE FRACTIONS. 53 6. Reduce,, 1 and I, to fractions having 30 for their common denominator. 15 10 7' 6 5 Ans. -, -0 30 3 and 30 30 ~_o 30' REDUCTION OF DENOMINATE FRACTIONS. 30. A denominate fraction is a fraction of a number of a particular denomination. Thus, 7 of a foot, 4 of a yard, I of a dollar, and 5 of a shilling are denominate fradtions. Reduction of denominate fractions is the changing of them from one denomination to another, without altering' their values. 31. Suppose we wish to reduce 3 of a pound sterling to an equivalent fraction of a farthing, we proceed as follows: Since there are 20 shillings in 1 pound, it follows that 3 J of a pound is the same as 20 times 3 - of a shilling, and this is also the same as 12 times 20 times 3 of a penny; which, in turn, is 4 times 12 times 20 times 3 of a farthing. Hence. An ( of a pound sterling is equivalent to 3 of 2 of -2 of A of a farthing. Again, let us reduce 3 of a farthing to an equivalent fraction of a pound sterling. In this case, we must use the reciprocals of - -, 1-, X; we thus find that 3 of a farthing is equivalent to 3 of I of 1 —, of J- of a pound sterling. 5* HIGHER ARITHMETIC. Hence, to reduce fractions of one denominate value to equivalent fractions of other denominate values, we have this RULE. I. When the given fraction is to be reduced to a higher denomination, multiply it by a compound fraction whose terms are the reciprocals of the successive denominate values, included between the denomination of the given fraction, and the one to which it is to be reduced. II. When the given fraction is to be reduced to a lower denomination, then multiply it by a compound fraction whose terms have units for their denominators, and for numerators the successive denominate values included between the denomination of the givenfraction and the one to which it is to be reduced. EXAMPLES. 1. Reduce a of an inch to the fraction of a mile. In this example, the different denominate values between an inch and a mile are 12 inches, 16l== — feet, 40 rods, and 8 furlongs;.. our compound fraction is - of 2 of I of ~; which, multiplied by the given fraction, produces 3 of % of 32 of of 4; canceling the 3 and 2 of the numerators, against a part of the 12 of the denominators, we get, 8 X 3340 8- 168960' 2 Therefore, A of an inch is equivalent to 6 896 of a mile. REDUCTION OF DENOMINATE FRACTIONS. 55 2. Reduce l-i-3-w of a solar day to an equivalent fraction of a second. In this example, the successive denominate values between a solar day and a second, are 24 hours, 60 minutes, and 60 seconds; therefore, our compound fraction is ~4 of a of rA~; which, multiplied by the given fraction, becomes __ of 2 of 0- of -1Q; this becomes, after canceling like factors, 4 of a second. 3. Reduce. 6 4 of a yard to the fraction of a mile. Ans. -00 4. Reduce 27 of-a gill to the fraction of a gallon. Ans. 96 -. 5. Reduce -3-o of a pound to the fraction of a ton. Ans..1-3 6. Reduce ~ of a mile to feet. Ans. 1760 feet. 7. Reduce - of - of - of a yard to the fraction of a mile. Ans. -6 9 60. 8. Reduce 1 of - of 24 of a gallon to the fraction of a gill. Ans..4 9. Reduce 2 of 4 of a hogshead of wine to the fraction of a gill. Ans. - 2=5971 gills. 10. Reduce ~ of of 4 yards to the fraction of an inch. Ans. -24P=34{ inches. 11. Reduce 4 of -3 of a farthing to the fraction of a shilling. Ans. 4-d-. 12. Reduce -7-9 of an ounce to the fraction of a pound avoirdupois. Ans. 9 4 4. 32. To find what fractional part one quantity is of another of the same kind, but of different denominations. Suppose we wish to know what part of 1 yard 2 feet 56 HIGHER ARITHMETIC. 3 inches is; we reduce 1 yard to inches, which gives I yard=36 inches; we also reduce 2 feet 3 inches to inches, which gives 2 feet 3 inches=27 inches. Now it is obvious that 2 feet 3 inches is the same part of one yard that 27 is of 36, which is j-z=a-. Hence, we deduce this RULE. Reduce the given quantities to the lowest d&nomination mentioned in either; then divide the number, which is to become the fractional part, by the other number. EXAMPLES. I. What part of ~3 4 s. I d. is 2 s. 6 d.? In this example, the quantities, when reduced, become ~3 4 s. I d.=769 d.; and 2 s. 6 d.=30 d.; therefore 6-0e is the fractional part which 2 s. 6 d. is of ~3 4s. Id. 2. What part of 3 miles 40 rods is 27 feet 9 inches? Ans. 2-~. S. What part of a day is 17 minutes 4 seconds?:Ans. 675 4. What part of $700 is $5.30? Ans. 5fo 5. What fractional part of 2 hogsheads is 3 pints? Ans. 3. 6. What part of $3 is 22 cents r Ans.,-. 7. What part of 10 shillings 8 pence is 3 shillings I penny? Ans..8. What part of 100 acres is 63 acres, 2 roods, and 7 rods of land? Ans. +,-.-. REDUCTION OF DENOMINATE NUMBERS. 57 33. To reduce a fraction of any given denomination to whole denominate numbers. Suppose we wish to know the value of a of a yard; we know that a of a yard equals - of + of a quarter= — of a quarter= —1 quarter -+ { of a quarter. Again, - of a quarter equals ~ of - of a nail=-2 nails. Therefore, ~ of a yard equals 1 quarter 2 nails. Hence, we deduce this RULE. Multiply the numerator by the units in the next inferior denominate value, and divide the product by the denominator; multiply the remainder, if any, by the next lower denominate value, and again divide the product by the denominator; continue this process until there is no remainder, or until we reach the lowest denominate value. The successive quotients will form the successive denominate values. EXAMPLES. 1. What is the value of -N- of an hour? In this example, - of an hour=-* of So of a minute =12 minutes. 2. What is the value of 3 of 1 yard? Ans. 1 quarter, 27 nails. 3. What is the value of ~ of ~ of one mile? Ans. 1 furlong, 20 rods. 4. What is the value of 7 of A of 1 cwt.? Ans. 1 quarter, 12 pounds. 5. What is the value of - of 14 miles, 6 furlongs? Ans. 2 miles, 3 furlongs, 26 rods, 11 feet. 58 HIGHER ARITIIMETIC. 6. What is the value of 1 of 3 of 2 days of 24 hours each? Ans. 9 hours, 36 minutes. 7. What is the value of 1 of -5 of -ih- of an hour? Ans. 5 minutes, 37~ seconds. 8. What is the value of - of - of 9 hours and 18 minutes? Ans. I hour, 33 minutes. DECIMAL FRACTIONS. 69 CHAPTER III. DECIMAL FRACTIONS.'34. A DECIMAL FRACTION is that particular form of a vulgar fraction whose denominator consists of a unit followed by one or more ciphers. Thus, -, -3-,1 -o o, and 303, are decimal fractions. In practice, the denominators of decimal fractions are not written, but they are always understood. Thus, instead of, - 37- and 1,X, we write 0'3, 0'07, 0'037, and 0'0001. The first figure on the right of the period, or decimal point, is said to be in the place of tenths, the second figure is said to be in the place of hundredths, the third in the place of thousandths, and so on, decreasing from the left towards the right, in a ten-fold ratio, the same as in whole numbers. The following table will exhibit this more clearly. NUMERATION TABLE OF WHOLE NUMBERS AND DECIMALS. *. — 2 ASCENDING. DESCENDING. This table is in accordance with the French method of number. ing, where each period of three figures changes its denominate values 613 aC1.. CcS m~cC*2CCCl C 0cC a CCC~ C'E C2a ~ 2C C1 22 2CC2 a ~ 33 3 333.33333 ASEDIG..r O~ DECNIG * This table isin accordance wth the French mthod of number ug, wher eac perio of thre fiue chne it eoiae value 60 HIGHER ARITHMETICS EXAMPLES. I. Write 7 tenths; 365 thousandths; 75 millionths. Ans. 0'7; 0'365; 0'000075. 2 Write 37 hundredths; 5 tenths; 3781 ten millionths. Ans. 0'37; 05; 0'0003781. 3. Write 43 hundredths; 3456 ten thousandths. Ans, 0"43; 0'3456. 4. Write 13 billionths; 3 ten billionths. Ans. 0'000000013; 0'0000000003. 35. Since decimals, like whole numbers, decrease from the left towards the right, in a ten-fold ratio, they may, when connected together by means of the decimal point, be operated upon by precisely the same rules as for whole numbers, provided we are careful to keep the decimal point always in the right place. Annexing a cipher to a decimal, does not change its value. Thus, 0 3=0 3030300=, &c. But prefixing a cipher is the same as removing the decimal figures one place further to the right, and, therefore, each cipher thus prefixed reduces the value in a ten-fold ratio. Thus, 0'3 is ten times 0'03, or a hundred times 0'003. ADDITION OF DECIMALS. 36. FROM what has been said under Art. 35, we deduce the following RULE. Place the numbers so that the decimal points shall be directly over each other, and then add as in whole numbers. ADDITION OF DECIMALS. 61 EXAMPLES. 1. Find the sum of 47'3; 37'672; 1'789101; 88'9134, and 0'0037. Operation. 47'3 37'672 1'789101 88'9134 0'0037 Ans. 175'678201. 2. What is the sum of 0'67; 0'0371; 1100'0001; 47'5; 29'0037; 1'000005, and 33'033? Ans. 1211'243905. 3. What is the sum of 1'8; 40'06; 120'365; 47'003; 1100'0001; 31'11101, and 3'0001? Ans. 1343'33921. 4. What is the sum of 13'29; 14'2835; 111'117; 4'006; 67'88864, and 496'446? Ans. 707'03114. 5. What is the sum of 37'345; 8'26; 19'0005; 7'534; 10'94, and 103'729? Ans. 186'8085. 6. What is the sum of 0'90058; 7'634; 3'007956, and 1'1? Ans. 12'642536. 7. What is the sum of 47'635; 3'13; 0'003001; 4'5787; 0'40005, and 4112'3789? Ans. 4168'125651. 8. What is the sum of 17'154; 32'004501; 49'345; 6'4, and 1'0005? Ans. 105'904001..9. What is the sum of 4'996; 38'37; 421'633; 5'65, and 4'29? Ans. 474'939. 10. What is the-sum of 57'41; 365'0001, and 1'101? Ans. 423'5111. 6 62 HIGHER ARITHMETIC. 11. What is the sum of 2'4999; 47'121212; O'1, and 411'001? Ans. 460'722112. 12. What is the sum of 433'9; 777'5; 67'06, and 35'88? Ans. 1314'34. SUBTRACTION OF DECIMALS. 37. FROM what has been said under Art. 35, we infer the following RULE. Place the smaller number under the larger, so that the decimal point of the one may be directly under that of the other. Then proceed as in subtraction of whole numbers. EXAMPLES. 1. From 213'5734 subtract 87'657237. Operation. 213'5734 87'657237 Ans. 125-916163. 2. From 385'76943 subtract 72'57. Ans. 313'19943. 3. From 0'975 subtract 0'483764. Ans. 0'491236. 4. From 0'5 subtract 0'0003. Ans. 0'4997. 5. From 96'5 subtract 0'000783. Ans. 96'499217. 6. From 23'005 subtract 13'000378. Ans. 10'004622. MUtTIPLICATION OF DECIMALS. 63 7. From 110'001001 subtract 11'010002. Ans. 98'990999. MULTIPLICATION OF DECIMALS. 38. LET US multiply 0'47 by 0'6. If we put these decimals in the form of vulgar fractions, they will become 1-4-07, and,!; these, multiplied by Rule under Art. 25, give 1407 X 6 —8 Now it is obvious that there will be, in all cases, as many ciphers in the denominator of the product as there are in the denominators of both the factors added together. Hence, the following RULE. Multiply the two factors after the same manner as in whole numbers; then point off, from the right of the product, as many figures for decimals as there are decimal places in both the factors. If there are not so many places of figures, supply the deficiency by prefixing ciphers. EXAMPLES. 1. Multiply 3'753 by 1'656. Operation. 3'753 1'656 22518 18765 22518 3753 Ans. 6'214968. 64 HIGHER ARITHMETIC. 2. What is the product of 0'005 into 0'017? Ans. 0'000085. 3. What is the product of 0'376 into 0'0076894? Ans. 0'0028912144. 4. What is the product of 0'576 into 0'3854? Ans. 0'2219904. 5. What is the product of 0'43 into 0'65? Ans. 0'2795. 6. What is the product of 3'9765 into 4'378? Ans. 17'409117. 7. What is the product of 415'314 into 7'3004? Ans. 3031'9583256. 8. What is the product of 7'42 into 11'1415? Ans. 82'66993. ABRIDGED MULTIPLICATION OF DECIMALS. 39. ABRIDGED multiplication may be advantageously employed, when one or both of the factors are expressed approximately in decimals. Suppose'we wish the product of I and.; if we -I-18 employ the rule under Art. 25, we shall find x In decimals, we have ~=0 33333, &c.; 6=0'16666, &c., and 1-8=0'05555, &c. We will now multiply together the decimal values of 3, and a, employing, in the first operation, 3 decimal places in each factor; 4 places in the second operation, and 5 in the third, as follows: ABRIDGED MUITIPLICATION OF DECIMALS. 65 F.irst Operation. Second Operation. Th7rd Operation. 0'333 0'3333 0'33333 0'166 0 1666 0'16666 1 998 1 9998 1 99998 19 98 19 998 19 9998 33 3 199 98 199 998 0-055,278 333 3 1999 98 0'0555,2778 3333 3 0'05555,27778 In the first operation, the result is true to only 3 places of decimals; in the second, it is true to 4, and in the third, to 5; and were we to employ a greater number of decimals in each factor, the product would be found to be accurate to only as many decimal places as there were in each factor. And in all cases, when the factors are approximate decimals, the whole number of decimals obtained in the product, by the usual method of multiplication, is not accurate. By the following rule, we may very much abridge the labor of multiplying, and still obtain the product with the same degree of accuracy as by the usual rule. Our rule for contracting the work of multiplying decimals, is as follows: RULE. I. Multiply the multiplicand by the left-hand figure of the multiplier. II Multiply the multiplicand, deprived of its righthand figure, by the second figure of the multiplier, counting from the left. III. Multiply the multiplicand, deprived of its two 6* 66 IHIGHER ARITIHMETIC. right-hand figures, by the thirdfigure of the multiplier, counting front the left. Continue this process until all the figures of the multiplier have been used. Observe to place the successive products so that their right-hand figures shall be directly under each other. NOTE. —In omitting successively the different figures on the right of the multiplicand, we must so far use them as to determine what there would be to carry into the next column. The student may, perhaps, find some difficulty in fixing the decimal point in the right place. Whenever he is at a loss in this respect, he can multiply a few of the left-hand figures of each factor by the common method, by which means he will be enabled to determine the true place for the decimal point. Or, which perhaps would be more simple, let the decimal point be fixed in the first partial product, which may be done by the usual rules for decimals. EXAMPLES. 1. Multiply 0'37894 by 0'67452. Operation. 0'37894 0'67452 0'227364 26526 1516 189 Ans. 0'255602 Explanation. First. We multiply the multiplicand 0'37894 by 6, the left-hand figure of the multiplier, which gives the first partial product, 227364 ABRIDGED MULTIPLICATION OF DECIMALS. 67 Secondly. We multiply 0'3789, which is the multiplicand deprived of its right-hand figure, by 7, the second figure of the multiplier, observing to carry 3, since the figure cut off, multiplied by 7, gives 28, which is nearer 30 than 20; we thus obtain 26526 for the second partial product. Thirdly. Multiplying 0'378 by 4, observing to carry 4, we obtain 1516 for the third partial product. Fourthly. Multiplying 0'37 by 5, observing to carry 4, we obtain 189 for the fourth partial product. Fifthly. Multiplying 0'3 by 2, observing to carry 1, we get 7 for the fifth partial product. As a second example, we will find the product of 3 by o, using in the first operation, 3 decimal places in each factor; in the second operation, we will use 4, and in the third, we will use 5, as follows: First Operation. Second Operation. Third Operation. 0'333 0'3333 0'33333 0 166 0'1666 0'16666 0'0333 0'03333 0'033333 200 2000 20000 20 200 2000 0'0553 20 200 0'05553 20 0'055553 From the above work, it will be seen that the results of these three operations have the same degree of accuracy as when performed by the usual rule. 3. Multiply 0'3785 by 0'4673. 68 IIGHIER ARITHMETIC C Operation. 0'3785 0'4673 0'15140 2271 265 11 Ans. 0'17687 4. Multiply 0'00524486 by 0'99993682. Operatio 0'00524486 0'99993682 0'004720374 472037 47204 4720 157 31 4 Ans. 0'005244527 5. Multiply 108'2808251671 by 1'9614591767. Operation. 108'2808251671 1'9614591767 108'2808251671 97 4527426504 6 4968495100 1082808252 433123301 54140412 9745274 108281 75796 6497 758 Ans. 212'3884181846 DIVISION OF DECIMALS. 69 6. Multiply 0'009416517988 by 0'999936883996. Ans. 0'0094159236548. 7. Multiply 0'0000375229 by 0'0000275177. Ans. 0'000000001032543. 8. Multiply 0'999936883996 by 0'999955663612. Ans. 0'9998925504063. 9. Multiply 0'587401052 by 0'018468950. Ans. 0'0108486807. 10. Multiply 91'6264232009 by 0'0172021234. Ans. 1'576169038601. 11. Multiply 212'3880258928 into itself. Ans. 45108'67354264. DIVISION OF DECIMALS. 40. IN multiplication, we have seen that there are as many decimal places in the product as there are in both the factors; and, since division is the reverse of multiplication, it follows that the number of decimal places in the quotient must equal the excess of those in the dividend, above those of the divisor. Hence, to divide one decimal expression by another, we have this RULE. Divide as in whole numbers, and point off as many places from the right of the quotient, for decimals, as the decimal places in the dividend exceed those of the divisor. If there are not as many figures in the quotient as this excess, supply the deficiency by prefixing ciphers. 70 HIGIIER ARITHMETIC. EXAMPLES. 1. Divide 3'475 by 4'789. Operation. 4'789)3'475000(0'725 3 3523 12270 9578 26920 23945 2975 In this example, the number of decimal places in the dividend, including the ciphers which were annexed, is 6, whilst the number of places in the divisor is 3; therefore, we make 3 places of decimals in the quotient. We might continue to annex ciphers to the remainder, and thus obtain additional decimal figures. 2. What is the quotient of 78'56453 divided by 4'78? Ans. 16'436. 3. What is the quotient of 1561'275 divided by 24'3? Ans. 64'25. 4. What is the quotient of 0'264 divided by 02? Ans. 1'32. 5. What is the quotient of 3'52275 divided by 3'355? Ans. 1'05. 6. What is the quotient of 901'125 divided by 2'25? Ans. 400'5. ABRIDGED DIVISION OF DECIMALS. 41. IF we divide 0'30679006 by 0'27610603, by the last rule, our work will be as follows: ABRIDGED DIVISION OF DECIMALS. 71 Operation. 0'27610603)0'30679006(1 1111313 27610603 3068403 0 2761060 3 307342 70 276106 03 312361670 27610 603 3626 0670 2761 0603 865 00670 828 31809 361688610 27 610603 9 078070 8 2831809 17948261 By simply inspecting the above work, it is obvious that all that part of the work which is on the right of the vertical line can in no way affect the accuracy of our quotient figures. By the following rule, we may perform the work of division so as to exclude all that part of the work on the right of the vertical line, thereby shortening the work, and still obtaining as accurate a result as by the last rule. To contract the work in the division of decimals, we have this RULE. Proceed as in the last rule, until we reach that point of the work where it would be necessary to annex ciphers to the remainder. Then, instead of annexing a cipher to the remainder, omit the right-hand figure of the divisor, and we shall obtain the next figure of the quotient; 72 HIGHERL ARITHIMETIU. and thus continue, at each successive figure of the quotient, to omit the right-hand figure of the divisor, until there is but one figure in the remainder. NOTE.-If we regard the dividend as the numerator of a fraction whose denominator is the divisor, the quotient will be the value of such fraction. Annexing a cipher to the numerator of a fraction is equivalent to multiplying its value by 10, and omitting the righthand figure of the denominator is also equivalent to multiplying the value of the fraction by 10. Hence, in the operation of division of decimals, instead of annexing a cipher to the dividend, as in the ordinary rules, we may, instead thereof, omit the right-hand figure of the divisor, as in the foregoing rule. EXAMPLES. 1. What is the quotient of 365'424907 divided by 0'263803? Operation. 0'263803)365'424907(1385'21892 263 803 101 6219 79 1409 22 48100 21 10424 1 376767 1 319015 57752 52761 4991 2638 2353 2110 243 237 6 5 1 ABRIDGED DIVISION OF DECIMALS. 73 2. What is the quotient of 0'123456 divided by 1 912478? Operation. 1'912478)0' 123456(0'064552 114748 8708 7650 1058 956 102 96 6 4 2 3. What is the quotient of 0'52600000 divided by 0'5260202? Operation. 0'5260202)0'52600000(0'9999616 47341818 5258182 4734182 524000 473418 50582 47342 3240 3156 84 53 31 31 4. What is the quotient of 7'45678 divided by 4'56789? Ans. 1'63243, 7 74 HIGHER ARITHMETIC. 5. What is the quotient of 7'632038 divided by 3'716048? Ans. 2053805. 6. What is the quotient of 2 divided by 15'314865? Ans. 0'13059207. 7. What is the quotient of 0'926954 divided by 0'3547898? Ans. 2'612685. 8. What is the quotient of 13'75892 divided by 6'76897? Ans. 2'03264. 42. To change a vulgar fraction into an equivalent decimal fraction. It is obvious that the rule under Art. 33 will apply to this case by considering all the denominate values as decreasing regularly in a ten-fold ratio. Hence, this R ITLE. Annex a cipher to the numerator, and then divide by the denominator; to the remainder annex another cipher, and again divide by the denominator, and so continue, until there is no remainder, or until we have obtained as many decimal figures as may be desired. The successive quotients will be the successive decimal figures required. EXAMPLES. 1. What decimal fraction is equivalent to'U? Operation. 16)100(0'0625 96 40 32 80 0 DECIMATION OF VULGAR FRACTIONS. 75 2. What decimal is equivalent to -?? Ans. 0'05555, &c. 3. What decimal is equivalent to -2? Ans. 0'05. 4. What decimal is equivalent to -x? Ans. 0'04. 5. What decimal is equivalent to 3? Ans. 0'3333, &c. 6. What decimal is equivalent to TT? Ans. 0'048. 7. What decimal is equivalent to 7? Ans. 0'875. 8. What decimal is equivalent to 3? Ans. 0'75. Since, in the above process of decimating a vulgar fraction, each successive dividend terminates with a zero, it follows that the right-hand figure of the remainder may be found by multiplying the right-hand figure of the denominator of the vulgar fraction by the quotient figure, and subtracting the right-hand figure of the product from 10; or, which is the same thing, if we subtract the righthand figure of the denominator from 10, and multiply the remainder by any decimal figure, the right-hand figure of the product will be the same as the right-hand figure of the remainder. 43. It will often happen, as in examples 2 and 5, of the last article, that the process will never terminate, in which case there is no decimal value which is accurately equal to the vulgar fraction. Since we constantly multiplied the remainders by 10, it follows that whenever the denominator of the vulgar fraction contains no prime factors different from those which compose 10, viz., 2 and 5, then the decimal value 76 HIGHER ARITHMETIC. will terminate. But, in all other cases, the decimal expression must consist of an infinite number of figures. Hence, to determine whether a given vulgar fraction can be accurately expressed in decimals, we have this RULE. Decompose the denominator of the vulgar fraction, when reduced to its lowest terms, into its prime factors, (by Rule under Art. 7,) then, if there are no prime factors different from 2 and 5, the vulgar fraction can be accurately expressed by decimals; but if it contain dif ferent factors, it cannot be expressed in decimals. EXAMPLES. 1. Can the vulgar fraction 3 be accurately expressed in decimals? In this example, we find that 386 =2 x 193; so that the denominator contains the prime factor, 193, which is different from 2 or 5; consequently,.33 cannot be accurately expressed in decimals. 2. Can the vulgar fraction — 7- be accurately expressed in decimals? Ans. It cannot. 3. Can the vulgar fractions, having for denominators 640, be expressed in decimals accurately? Ans. They can. 44. When a vulgar fraction can be accurately expressed in decimals, we may determine the number of decimal places by the following DECIMATION OF VULGAR FRACTIONS. 77 RULE. Decompose the denominator, after the fraction is reduced to its lowest terms, into its prime factors, (by rule under Art. 7,) which factors cannot differ from 2 and 5, (by rule under Art. 43.) The highest exponent of 2, or 5, will be the number of decimal places sought. EXAMPLES. 1. How many places of decimals will be required to express -j? In this example, we find 40 23 x 5, where the highest exponent is 3; therefore, the number of decimal places is 3. 2. How many places of decimals will be required to express T- 7? Ans. 3. 3. How many places of decimals will be required to express 3 Ans. 5. 4. How many places of decimals will be required to express 6? Ans. 4. 5. How many places of decimals will be required to express -37? Ans. 5. 6. How many places of decimals will be required to express 43? Ans. 6. 45. When many figures in the decimal are required, we may proceed as follows: Required the decimal value of 2? Following the rule under Art. 42, we get this 7* 78 HIGHIER ARITHMETIC. Operation. 29)100(0'03448 87 130 116 140 116 240 232 8 We have continued this process until we have found a remainder consisting of but one figure; placing this remainder, when divided by 29, at the right of the quotient, agreeably to the usual rules of division, we get, I. -' —=00344S8-A,. Multiplying this by 8, we get -2=0'27586 — 9. Substituting this value of n,- in I., we get, II. -9 —=0'0344827586~-; this, multiplied by 6, gives - =0'2068965517{ —; which, substituted in II., gives, III. 2- =0'03448275862068965517-g7. Again multiplying by 7, we get -7 —=0.241379310344827586202. Substituting this in IIT., we get, IV. -- = 003448275862068965517241379310344827586202. In the expression 2 — 0 03448-5,, the numerator 8, of the vulgar fraction -89, is the fifth remainder; and in the expression -j=0'0344827586-w, the numerator 6, of the fraction n6,-, is the tenth remainder; but this remainder 6, was obtained by multiplying the 5th remainder, which is 8, into itself, and dividing the product, 64, by 29, we thus found the remainder, 6. Agcain, the 20th remainder, DECIMATION OF VULGAR FRACTIONS. 79 which is 7, was found by multiplying the 10th remainder into itself, and dividing the product by 29. For a similar reason, the remainder of the product of any two remainders will give the remainder corresponding with the sum of the numbers denoting their order; thus, the 5th multiplied by the 7th, will give the 12th; the 6th multiplied by the 9th will give the 15th, and so on for other combinations. 46. There is another way of decimating, which is as follows: Decimate of According to rule under Art. 42, we find, 97)100(0-01 97 To continue this process, we must add ciphers to this remainder in the same way as we did to the numerator, 1. Now, the remainder being 3 times as large as the first numerator, it follows that the next two decimal figures must be 3 times the two just obtained, that is, 3 x 01- 03; and, for a similar reason, we must multiply 03 by 3, to obtain the next two figures, and so on. Proceeding in this way, we find, - = 0'0103092781 243 729 2187 6561 &c. = 0'010309278350515, &c. 80 HIGHER ARITHMETIC. Decimating 4 by the above plan, we get =0-248 32 64 128 256 &c. =0 0249999, &c., which will constantly approximate towards 0'25; hence, 4 =0'25. Decimating I by this method, we get, - =0'1248 16 32 64 128 256 &c. 8-=01249999, &c., which will constantly approximate towards 0'125. 47. When the decimal figures, obtained by converting a vulgar fraction into decimals, do not terminate, they must recur in periods, whose number of terms cannot exceed the number of units in the denominator, less one. For, all the different remainders which occur must be less than the denominator; and, therefore, their number cannot exceed the denominator, less one; and, whenever we obtain a remainder like one that has previously occurred, then the decimal figures will begin to repeat. Decimals which recur in this way are called repetends. COMPOUND REPETENDS. 81 When the period begins with the first decimal figure, it is called a simple repetend. But when other decimal figures occur before the period commences, it is called a compound rep etend. A repetend is distinguished from ordinary decimals by a period, or dot, placed over the first and last figure of the circulating period. 48. The following vulgar fractions give simple repetends: =0.3. 7 =0142857. =0oi. 1T= 009. I-= —0'076923. T1= 0'05882352941 17647. T= 0'052631578947368421. - =-0'047619. -= 0'04347826086956521739613. 49. The following ones give compound repetends. =0 16. 1- =0' 083. -1 -= 007] 4285. -1 =-006. 18 Y1= -0.045. -1 =0 041 6. 82 HGHIERI ARITHMETIC, 50. Whenever the prime factors of the denominator of a vulgar fraction contain neither of the factors 2 and 5, the repetend will be simple. But when they contain one or both of the factors 2 and 5, together with other factors, then the repetend will be compound. 51. Those simple repetends which have as many terms, less one, as there are units in the denominators of their equivalent vulgar fractions, we shall call perfect repetends. The following are all of the perfect repetends, whose denominators are less than 100. 0 =0'142857. _ = 0'0588235294117647. Ilg = 0'052631578947368421. —' = 0'0434782608695652173913.'1 = 0'03448275862068965517241 37931. _ = 0'62127659574468085106382 97872340425531914893617o _ s 0'01694915254237288135593220338 98305084745762711864406779661.i _- = I 0' 016393442622950819672131147540 983606557377049180327868852459. 0'01030927835051546391752577319587 1T~= g 62886597938144329896907216494845 l 36082474226804123711340206185567. PERFECT REPETENDS. 83 The value of T may be made to assume the following forms: -17 = o0'0 = 0.05 1} O- =o058s1 _ 0'o 588s-s=O- 0o5882s = 0'058823 = 00588235 0'05882352 &c., where each successive value is extended one decimal place further than its preceding value. The numerators of the vulgar fractions connected with the above decimal expressions are the successive remainders found in the operation of converting the vulgar fraction -17 into a decimal. (Art. 42.) If this process of decimating be continued, it will be found to give a simple repetend, consisting of 16 places of figures; it is, therefore, a perfect repetend. We will arrange this repetend by placing above each figure its corresponding remainder, as follows; 10 15 14 4 6 9 5 16 7 2 3 13 11 8 12 10 15 14,&. = 0 5882352941 1 7 6 4 7 0 5 8, c. If we fix our attention upon a particular remainder, as the fifth, for instance, which is 6, it is evident that the decimals which follow, as 3529, &c., continued to infinity, must express the decimal value of 167; for, had we terminated our division after the fifth decimal figure was obtained, we should have had -T = 0'05882-, where 6 stands instead of the decimal figures which follow 005882, so that the decimal figures following the remainder, 6, is equal to ~,6 In the same way, the decimals which follow any other remainder is the value of the vulgar fraction whose denominator is 17, and whose numerator is said remainder. We have already said that the decimal figures would commence repeating when a remainder is found like one which has previously occurred, (Art. 47.) A perfect 84 IIGIIE ARITIMETIC. repetend has been defined (Art. 51,) as one whose number of decimal figures is equal to one less than the units in the denominator of the vulgar fraction from which it is derived. Therefore, in converting the vulgar fraction I into a perfect repetend, every number, from 1 to 16, inclusive, must appear as a remainder. Let us suppose we have reached that point in the process of decimating, which gives 16 for the remainder; then the decimals which follow, being the value of 1 6, must, when added to the preceding decimals, the value of -, make a suc(ession of 9's, as 99999, &c., since - +~=l, 1 which, expressed in decimals, is 0'99999, &c., continued to infinity. Hence, when we have obtained as many figures beyond the remainder 16, as we had before we found this remainder, the decimal figures will begin to repeat. But lo, giving a perfect repetend, must extend to 16 figures before repeating; consequently, 16 will occur as the 8th remainder, or when we have obtained one half the number of decimals in the period, and such must be the case with all perfect repetends. Therefore, the decimal figures of the first half of the period of a perfect repetend, being added to the figures of the second half, must give 99999, &c.; which, if considered as a decimal, is equivalent to a unit. Hence, also, the remainders of the first half of the period, being added to the remainders of the second half, must make, respectively, 17, since their corresponding decimal values make a unit, which is equivalent to If. Whenever the number of figures in the period of a simple repetend, arising from decimating a vulgar fraction, whose numerator is 1, and denominator a prime, is even, the remainder which occurs at the middle of the COMPLEMIENTARY REPETENDS. 85 period will be one less than the denominator of the equivalent vulgar fraction, and the figures of the first half of the period, added to those of the second half, will give 99999, &c.; and their corresponding remainders added, must give the denominator of the equivalent vulgar fraclion. Such repetends may be called COMPLEMENTARY REPETENDS. They, of course, include all perfect repetends, as well as many which are not perfect. The following complementary repetends are not perfect. 10 1 T1v =0'0 9. 10 9 I2 3 4 1 1 =0'0769 23. 10 27 51 72 63 46 22 1 - =0-0 1 3 6 9 8 6 3. 10 11 21 32 53 85 49 45 5 50 55 16 71 87 69 67 47 25 72 8 80 88 Half a =0'0 1 1 2 3 5 9 55 0 5 6 1 7 9 7 7 5 2 8 0 8. period. 10 100 91 1 T- = 0 0 0 9 9. 10 100 73 9 90 76 39 81 89 66 42 8 80 79 69 72 102 H = 0'0 0 9 7083 786 4 7 7 676. Hfaperiod. The following vulgar fractions, when decimated, give an even number of figures in a period, and still they are not complementary repetends, their denominators not being primes. -' =0'047619.,=.6. 3. -" =0'025641. F V=00'204081, &c., to 42 places. -1 = 0o0196078431372549. 8 St, IHIGnIER ARITHIMETIC. 52. PERFECT REPETENDS possess somle very remarkable properties, which we will explain by means of the following figure: In this figure, the inner circle of figures, commencing at the 0, directly under the asterisk, and counting towards the right hand, is the circulating period of JThe outer circle of figures, commencing at the same place, and counting in the same direction, are the successive remainders which will occur in the operation of decimating -j',, (by Rule under Art. 4.) In this circle of remainders, all the numbers from 1 to 28, inclusive, occur, but not in numerical order. From what has been said, we infer the following properties, which are common to all perfect repetends: PERFECT REPETENDS. 87 I. The sum of any two diametrically opposite figures of the circle of decimals, will be 9. II. The sum of any two diametrically opposite terms in the circle of remainders, will make the denominator 29. III. If we subtract the right-hand figure of the denominator from 10, and multiply the remainder by any decimal figure of the inner circle, the right-hand Jigure of the product will be the same as the right-hand figure of the corresponding remainder of the outer circle. IV. Commencing the circle of decimals at any point, and counting conLpletely round, it will be the perfect repetend-of the vulgar fraction, whose denominator is the same as in the first case, but whose numerator is the remainder in the outer circle, standing one place to the left. V. If we divide the product of any two remainders by 29, what remaing will be the remainder in the outer circle, corresponding with the place denoted by the sum of the places of the two numbers. From the IVth property, it follows that this same circle of decimals expresses the decimal value of all proper vulgar fractions, whose denominators are 29. The following figure, formed from the perfect repetend of the value of 2-, possesses similar properties to those just explained. Similar circles may be formed for all perfect repetends. 88 HIGHER ARITHMETIC. 53. If we take the perfect repetend arising from, we find =0' 42857. - 0 571428. 2 =0285714. 5 =0 714.85. =3-0.iS 71. 0 -,S57142. Taking the perfect repetend arising from ~~ we find 1~7 -0'588235294117647. 2 -' 176470588235294. -- 0O 1 764705882352941. -= 0'2352941176470588. -5 = 02941176470588235. -0 3529411764705882. yr = 04117647058823529. A8 = 0'i705882352941176. PERFECT REPETEN DS 89 Y-9= 0'5294117647058823. 1 - =0')5882352941176470. i =0'6470588235294117. 2=- 0'7058823529411764. 1 3 = 0'764705882352941 1. 14= 0 8235294117647058. T4 -=0'8823529411764705. 1-= 0'9411764705882352. We will arrange the complementary repetend arising from the vulgar fraction T I, in the form of a circle, as was done for perfect repetends, as follows: J/~~;l( 90 HIGHER ARITHMETIC. It will be seen that a complementary repetend possesses all the properties ascribed to the perfect repetend, as given under Art. 52, except the IV: 54. To change a decimal fraction into an equivalent vulgar fraction. Case I. When the number of places is finite, we can, from the definition of decimal fractions, Art. 34, deduce this R ULE. Make the given decimal the numerator of the vulgar fraction, and, for its denominator, write 1, with as many czihers annexed as there are decimal places. EXAMPLES. 1. What vulgar fraction is equivalent to the decimal 0'0625? ~6 2, or 625 0; this, reduced by Rule under Art. 17, gives,-',; therefore, 0'0625 —6,. 2. What vulgar fraction is equivalent to the decimal 0o134? Ans. I'3'-4 —3. What vulgar fraction is equivalent to the decimal 0'00125? Ans. 1 0o25o 0 -- 8 0 4. What vulgar fraction is equivalent to the decimal 0'0256? Ans.. 5. What vulgar fraction is equivalent to the decimal 0'06248? Anos. Ti`2 I_ 125o-o0 6. What vulgar fraction is equivalent to the decimal 0'001069? Ans. T d00-r-0 o. - REDUCTION OF REPETENDS. 91 Case II. When the decimal is a simple repetend. Since -=0'1, it follows that 02 must =-, 0'3=3, 04= 4, and so on; therefore, a simple repetend of one figure is equivalent to the vulgar fraction whose numerator is this figure, and whose denominator is 9. Again, — =0601; consequently, 0'07=, — 0'45=-4, and so on for other simple repetends of two places of figures. In a similar manner, we infer that 0A432 = 4.32 Therefore, we have the following RULE. Make the repetend the numerator; and, for the denominator, write as many nines as there are places of decimals. EXAMPLES. 1. What vulgar fraction is equivalent to 07'2? - 2; this, reduced by Rule under Art. 17, becomes 11. 2. What vulgar fraction is equivalent to 0123? Ans. 2-93 =41 3. What vulgar fraction is equivalent to the repetend 0'027? Ans. 27 = -. 4. What vulgar fraction is equivalent to the repetend 0 142857? Ans. g4 2 957_5. What vulgar fraction is equivalent to the repetend 0')01?3456i79? Ans. 9'934.597.999 -8 6. What vulgar fraction is equivalent to the repetend O0'123456789? Ans. 12.34789 1 3 7 17421.T- ( FT7WV- I T i T I 1,. 92 HIGHER ARITHMETIC. 7. What vulgar fraction is equivalent to the repetend 0oi2332i? Ans. 2 3 2 1 0 1 8. What is the value of 0'999 continued to infinity? Ans. 9-=1. 9. What is the value of 0'9876543206? Ans. 98765432080 [A very simple method of finding a vulgar fraction equivalent to any repetend, may be found in my ELEMENTS OF ALGEBRA.] Case III. When the decimal is a compound repetend. In this case, we obviously have the following RULE. I. Find the vulgar fraction which is equivalent to the decimal figures which precede those that circulate, by Rule under Case I. of this article. II. Find the vulgar fraction which is equivalent to the circulating part of the decimal, by Rule under Case II of this article; to the denominator of this fraction annex as many ciphers as there are decimals which precede the circulating part of the repetend; then add these two fractions together. EXAMPLES. 1. What vulgar fraction is equivalent to the compound repetend 0'343? Ans.,3-00 + -3 - 3 ~ - 2. What vulgar fraction is equivalent to the compound repetend 0'08783? Ans. T, + 3 - 1-3 3. What vulgar fraction is equivalent to 0'083? Ans. 80 + 3 - = -I. REDUCTION OF REPETENDS. 93 4. What vulgar fraction is equivalent to the compound repetend 0'0357142S? Ans. 3 + -71428 o = 5. What vulgar fraction is equivalent to the compound repetend 0'0714285 Ans. -7194 8 5 =-I 6. What vulgar fraction is equivalent to the compound repetend 0'123456? Ans. 123+ 9 41 11 — 43'oIf we take the last example, which is 04123456, and multiply it by 1000000, it will become 123456'456. Again, if we multiply 0'123456 by 1000, it will become 123'456. The difference of these two results is 123456456- 123'456= 123333. Now, since 123456'456 was 1000000 times the decimal 04123456, while 123'456 was 1000 times the same decimal, it follows that 123333 is (1000000-1000) times its value; that is, 123333 is 999000 times the value of 0 1 23456; hence, 0 1 23456 = s1233339 - 4111 the same as already found. A similar VVVoW 90 0 F- 3330 0 0, process may be employed for changing any repetend into an equivalent vulgar fraction. 94 HIGHER ARITHMETIC. CHAPTER IV. CONTINUED FRACTIONS. 55. IF we divide both numerator and denominator of the fraction 35 by the numerator, we obtain, 351 1 965 2+263 351. Again, performing the like operation upon the fraction 2 3, we find 263 1 3 5 1 351 1+88 263; this value of 2 63 substituted in I., we get, 351 1 965 2 1 1+88 88 1 263. Again, we find 263=2+87 88; which, substituted in II., gives, 351 1 III. 965 2+ 1 1+1 2+87 87 1 88. Again, 8 = 1 87; this value substituted in III., we finally obtain, CONTINUED FRACTIONS. 95 351 1 IV. 965 2+ 1 1+1 2+1 1+1 87. By a similar process, we find that 157 1 972 6+1 5+1 4+1 3+1 2. Such fractions as the above are called continued fractions. In the last example, the parts 6, 5, ~, &c., are called the first, second, third, &c., partial fractions. It has been proposed, by some authors, to write continued fractions in the following way, which is more compact: Thus, the preceding fractions may be written, 351 1 1 1 1 1 965-2+1+2+1+87. 157 1 1 1 1 1 972 6+5+4+3+ 2. If we seek for the greatest common measure of the numerator and denominator of the first fraction 3 5, by Rule under Art. 9, we shall have the following 96 HIGHER ARITHMETIC. Operation. 351)965(2 702 263)351(1 263 88)263(2 176 87)88(1 87 1)87(87 87 0 Here we discover that the successive quotients are the same as the successive denominators of the partial fractions which compose the continued fraction already drawn from 3 51 Hence, to convert a vulgar fraction into a continued fraction, we have this RULE. Seek, by Rule under Art. 9, the greatest common measure of the numerator and denominator of the given fraction; the reciprocals of the successive quotients will form the partial fractions which constitute the con. tinued fraction required. EXAMPLES. 1. Convert into a continued fraction. CONTINUED FRACTIONS. 97 Operation. 251)764(3 753 11)251(22 22 3i 22 9)11(1 9 2)9(4 8 1)2(2 2 The partial fractions are 3, -2, -, D i; therefore, we shall have, 251 1 764 3+1 22+1 1+1 4+1 2. 2. What continued fraction is equivalent to fifI? Ards.1 3+1 7+1 1+1 2+1 4+1 5+1 1+1 2. 9 98 HIGHER ARITHMETIC. 3. What continued fraction is equivalent to - r s? Ans. 1+1 2+1 1+1 1+1 7+I 2. 4. What continued fraction is equivalent to -8? A 1 Ans. 3+1 1+1 2+1 4+1 5. 56. Let us now endeavor to reverse the foregoing process; that is, let us seek the vulgar fraction which is equivalent to a continued fraction. If we take the continued fraction1 2+1 4+1 5, and omit all but the first partial fraction, its value will become 2. Again, omitting all but the first and second partial fractions, we find 1 3. 2+ 1 7 CONTI:NUED FRACTIONS. 99 Again, including one more partial fraction, we obtain 1 13 2+1 30 3+1 4. When we include the whole, we find 1 68 2+1 157 3+1 4+1 5. Our successive values, obtained in this way, are 2, T, 133, and A-. These values may be derived in the following manner: Take the first partial fraction for the first value; multiply both numerator and denominator by the denominator of the next partial fraction, and we get -; if we increase this denominator by 1, it will give the second value, 7. Again, multiplying numerator and denominator by the denominator of the next partial fraction, we get -2; if we increase this numerator by the numerator of the last value, also increase the denominator by the denominator of the last value, we get 33, which is the third value. Again, multiplying both numerator and denominator of this value, by the denominator of the next partial fraction, and to the respective products add the numerator and denominator of the preceding value, we obtain the last value, -6 857. This last value is the true value of the continued fraction, whilst the other values are successive approximations. 100 HIGHER ARITHMETIC. From what has been said, we derive the following Rule for finding the vulgar fraction equivalent to a continued fraction: RULE. Consider the symbol o as a fraction; then write this symbol, and the first partial fraction, for the first two terms of the approximate values. Multiply the numerator and denominator of the second approximate value, by the denominator of the next partial fraction, and to the respective products add the numerator and denominator of the next preceding approximate value, and the result will be the succeeding approximate value. Thus continue to multiply the last approximate value by the denominator of the succeeding partial fraction, and to the products add the numerator and denominator of the preceding approximate value; the result will be the succeeding approximate value. EXAMPLES 1. What vulgar fraction is equivalent to the continued fraction 3+1 2+1 5+1 4+1 6? In this example, we find the successive approximate values to be o0 2, I 4,,, and 29,397 38 1,~ _5 96 an [V CONTINUED FRACTIONS. 101 2. What are the approximative values of the continued fraction 3+1 1+1 2+1 4+1 5? Ans. ~, I, I, T3 13 and 283. What are the approximative values of the continued fraction 2+1 2+1 2+1 2? Ans. o a 2,5 and'2 4. What are the approximative values of the continued fraction 1+1 2+1 3+1 4+1 5+1 6+1 7+1 8+1 9? 0 1 2 7 3 0 1 5 7 9 72 6 9 6 1 5 6 6 6 0 Ans.''' T WT 102 HIGHER ARITHMErTIC. 5. What are the approximative values of the continued fraction 1 9+1 8+1 7+1 6+i 5+1 4+1 3+1 2? 0 1 8 57 350 1807 7578 2 4541 Ts T9 08 3- 520o31 9310 10 T485 8 60F1 35- 3 NA 8 Ans.,56 6 6 0 57. We will now show the application of the fore- * going principles of continued fractions by the solution of several practical questions: 1. Express approximately the fractional part of 24 hours, by which the solar year of 365 days, 5 hours, 48 minutes, and 48 seconds, exceeds 365 days. 5 hours, 48 minutes, 48 seconds=20928 seconds. 24 hours=86400 seconds. Therefore, the true value of the fraction required is 9 8 o 9 Now, converting 459 into a continued fraction, by Rule under Art. 55, we get 109 1 450 4 +1 7+1 1+1 3+1 1+1 2; CONTINUED FRACTIONS. 103 and'this, re-converted into its approximative values (by Rule under Art. 56,) gives 0, 4 1 v 8 343 12- 11 9o The fraction I agrees with the correction introduced into the calendar by JULIUS CGESAR, by means of bissextile or leap-year. The fraction 8 is the correction used by the Persian astronomers, who add 8 days in every 33 years, by having 7 regular leap-years, and then deferring the eighth until 5 years. 2. The French metre is 39'371 inches. Required the approximnative ratio of the English foot to the metre. In this example, the true ratio is 132.0 Operating upon this fraction, as in the last example, we find some of the first approximate values to be -,,-343, 23, 32 I 0 5' Hence, the foot is to the metre as 3 to 10, nearly; a more correct ratio is as 32 to 105. 3. The old Winchester bushel contains 2150'42 cubic inches, and the new Imperial bushel contains 2218'198 cubic inches. Required some of the approximative ratios of these numbers. In this case, we find some of the approximations to be 1 32 95 127 be, 3 27 331 981 TT Hence, the Winchester bushel is to the Imperial bushel as 32 to 33, nearly. Now, since in a bushel there are 32 quarts, it follows that the Imperial is a Winchester quart larger than the Winchester bushel, nearly. 4. What are some of the approximative values of the ratio of the diameter of a circle to its circumference? If we take the value of the circumference of the circle, whose diameter is 1, to 10 decimals, we have the vulgar 104 HIGHER ARITHIMETIC. fraction,o0 olo ooooo, given to find its approximative values. Proceeding with this, as in the former examples, we find some of the first approximative values to be I, —, I o, -'3 &c 3 3 3 3 5 &C 58. Continued fractions have been the means of obtaining elegant approximations to the roots of surds. As an example, let it be required to find the square root of 2; or, what is the same thing, the ratio of the side of the square to its diagonal. I I In the first place, we obviously have = 2 — 1+ 1 Now, if we multiply the numerator and denominator of the fraction V2-1 by /2+ 1, it will become, V/2 —1 I 1 I 1/2~-1 12-1 2+ 1 Therefore, we have, a/21 1 1 + -/2 — 1 2+ Again, the fraction /2-1 becomes, as before, 2+ 2-1, and by thus continuing this process, we find - 1 I to equal the following infinite continued fraction: to2 CONTINUED FRACTIONS 105 1 1~1 2+1 2+1 2+1 2+, &c. Some of the first approximative values of this fraction are 2 5 12 29 70 1 &c. are 1, l, 3, 2, - 99 239 C. 59. We will conclude this subject by pointing out some of the many remarkable properties which the approximative values of continued fractions possess. We will refer to the values just obtained for the ratio of the side of a square to its diagonal. I. These values are alternately too small and too large. Thus, o, 2, 2, and 9o, are too small, while, I, A, and 2 6, are too large. II. Any of these values difer from the true value by a quantity which is less than the reciprocal of the square of its denominator. Thus, 2, which is the ratio much used by carpenters in cutting braces, differs from the true ratio by a quantity less than (,_)2 = _9. III. Any two consecutive terms of these approximate values, when reduced to a common denominator, will differ by a unit in their numerators. Thus, 5, and 12, when reduced to a common denominator, become 8159, and.849 IV. The numerator and denominator of all approxi 106 HIGHER AR ITIIMETIC. mative values of continued fractions are prime to each other; that is, they have no common measure. [For a more extended developement of the properties of Continued Fractions, see my Treatise on ALGEBRA.] LAMBERT'S METHOD OF DE-COMPOUNDING VULGAR FRACTIONS. 60. THERE is another method of de-compounding fractions, which was first given by the celebrated LAMBERT. As an example, we will resume our fraction 3 66 which may be made to take these forms: 351 2 x 351 702 965 - 263 1 1 263 965 2x965 2x965 2 x 965 2 2 965 Again, 263 3 x 263 789 965-176 1 1 176 965 3x965 3x965 3x965 3 3 965 And, 176 5x176 880 965- 85 1 1 85 965 5x 965 5 x 965 5 x 965 5 5 X 965 And, 85 11 x 85 935 965- 30 1 1 30 965 11 x 965 11 x 965 11 x 965 11 11 965 And, 30 32 x 30 960 965 —5 1 1 5 965 32 x 965 32 X 965 32 x 965 32- 32 X 965' And, 5 1 965 193 DE-COMPOSITION OF FRACTIONS. 107 Hence, by successive substitutions, we find, 351 1 1 1 1 965 2 2x3 2x3x5 2x3x5 x11 1 1 2x3x5x 11 x32- 2x 3 x 5x 11 x32x 193 Here we observe that the successive terms are alternately plus and minus. We also see that the successive factors of the different denominators may be found by continually dividing the denominator, 965, by the numerator, 351, and the successive remainders. To make this more clear, we will give the above method of division at full length. Operation. 351)965(2 702 263)965(3 789 176)965(5 880 85)965(11 935.30)965(32 960 5)965(193 965 As a second example, de-compound the fraction _ 2 o, which is nearly the ratio of the English foot to the French metre. 108 HIGHER ARITHMETIC. OpElation. 1200)3937(3 3600 337)3937(11 3707 230)3937(17 3910 27)3937(145 3915 22)3937(178 3916 21)3937(187 3927 10)3937(393 3930 7, &c. Hence, the value of 3 2 o f 1 1 1 1 1 3 3.1 3.11.17 3.11.17.145 3.11.17.145.178 1 1 3.11.17.145.178.187 3.11.17.145.178.187.393 &c. It is obvious that the terms of this series converge very rapidly. If we use only one term of the series, we have I for the ratio. If we use two terms, we find 3 3 for the ratio; and in this way, we may find the successive approximate ratios. If we endeavor to de-compound the fraction 4 83673 by this method, we shall find it equivalent to this series of fractions: 1 1 1 1 1 1 1.2 1.2.3 1.2.3.4 1.2.3.4.5 1.2.3.4.5.6 1 1 + 1.2.3.4.5.6.7 1.2.3.4.5.6.7.9 DE-COMPOSITION OF FRACTIONS. 109 It will be observed that the factors constituting the denominators of these fractions are the successive digits, except that in the last term the digit 8 does not appear. Now, instead of this last term, we may write these two terms 1 1 +1.2.3.4.5.6.7.8 1.2.3.4.5.6.7.8.9' which are obviously equivalent in value. This change being made, we have 28673 1 1 1 1 1 1 45360 1 1.2 1.2.3 1.2.3.4 1.2.3.4.5 1.2.3.4.5.6 1 1 1 1.2.3.4.5.6.7 1.2.3.4.5.6.7.8 1.2.3.4.5.6.7.8.9 If we de-compound the fraction 4 8 - 3- by this method, we shall find it equal to 1 1 1 1 1 1 2 2.4 2.4.6 2.4.6.8 2.4.6.8.10 2.4.6.8.10.12 10 110 HIGHER ARITHMETIC. CHAPTER V. RULE OF THREE. 61. THE quotient arising from dividing one quantity by another of the same kind, is called a ratio. Thus, the ratio of 12 to 3, is 12. 3-=1-=4. The ratio of 15 yards to 5 yards, is )-' —=3. So that the ratio of one number to another is nothing more than the value of a vulgar fraction, whose numerator is the first term, and denominator the last term. The ratio of 7 days to 5 days, is 7. The ratio of 34 hours to 72 hours, is 31 = 13 When there are four quantities, of which the ratio of the first to the second is the same as that of the third to the fourth, these four quantities are said to be in proportion. Thus, 4, 6, 8, and 12, are in proportion, since the ratio of 4 to 6, is the same as 8 to 12. That is, 4- 8 Hence, a proportion is nothing more than an equality of ratios. The usual method of denoting that four terms are in proportion, is by means of points, or dots. Thus, 4 6:: 8: 12; where two points are placed between the first and second terms, and also between the third and fourth, and four points are placed between the second and third; which is read, 4 is to 6 as 8 is to 12. RULE OF THIREE. 111 The first and fourth terms of a proportion are called the extremes. The second and third terms are called the means. The first and second constitute the first couplet. The third and fourth constitute the second couplet. The two terms of a couplet must be' of the same name, or kind; since two quantities of different kinds cannot have a ratio. There call be no ratio between yards and dollars; but the numbers which represent the number of yards and dollars may have a ratio. Since, in a proportion, the quotient of the first term, divided by the second, is equal to the quotient of the third, divided by the fourth, it follows that the product of the extremes is equal to the product of the means. Hence, if we divide the product of the means by the first term, we shall obtain the fourth term. This process of finding the fourth term by means of the other three terms, is called the Rule of Three. Suppose it is required to find what 156 yards of cloth will cost, if 22 yards cost $20'90. Had there been twice as many yards in 156 as in 22, they would obviously cost twice $20'90; were there three times as many yards, their cost would be three times $20'90, and so on for other ratios. Hence, the $20'90 must be repeated as many times as 22 yards is continued times in 156 yards; that is, -! —26 times. So that, if 22 yards of cloth cost $20'90, 156 yards will cost -'LP6L times $20'90== 71 times $20 90=78 times - of $20'90=78 times $1'90=$14&~20. Again, suppose 15 francs to equal $14 10, how many francs are there in $34'78? 112 HIGHER ARITIIMETIC. Since 15 francs is equal to $14'10, twice $14'10 would equal twice 15 francs; thrice $14 10 would equal thrice 15 francs, and so for other ratios. Therefore, 15 francs must be repeated as many times as $14'10 is contained times in $34'78; that is, 34 7 =- 3- times. Hence, the number of francs required is 37 of 15 francs=37 times x'l of 15 francs- 87 times 1 franc= 37 francs. The above method of working questions of the Rule of Three is contained in the following simple RULE. Of the three terms which are given, one will always be of the same kind as the answer sought; this will be the third term. Then if, by the nature of the question, the answer is required to be greater than the third term, divide the greater of the two remaining terms by the less, for a ratio; but?f the answer is required to be less than the third term, then divide the less of the two remaining terms by the greater, for a ratio. Having obtained the ratio, multiply the third term by it, and it will give the answer in the same denomination as the third term. NOTE.-Before obtaining the ratio by means of the first two terms, we must reduce them to like denominations. EXAMPLES. 1. If in 7 weeks there are 49 days, how many days are there in 21 weeks? In this example, the answer is required to be in days; therefore, we must take 49 days for our third term. And, since in 21 weeks there must be more days than in 7 weeks, we get our ratio by dividing 21 by 7, which RULE OF THREE. 113 gives A —; the third term, 49 days, multiplied by this, gives 49 days x —, which, by canceling, becomes 147 days, for our answer. 2. If a person perform a journey in 20 days, by traveling 10 hours each day, how long would it take him to perform the same if he travels 8 hours each day? In this example, our answer is required to be in days; therefore, we must take 20 days for our third term. And, since it will evidently take more days when he travels 8 hours each day, than it did when he traveled 10 hours each day, we must divide 10 hours by 8 hours, for our ratio, which becomes j-8; 20 days, multiplied by this, gives 20 days x-'-, which, by canceling, becomes 25 days for the answer. 3. If 1 of a pound of sugar cost 2 3 of a shilling, how much will A-A of a pound cost? In this example, our third term is -23 of a shilling. And, since -9 of a pound is less than'1, we must obtain our ratio by dividing A- by I-~, which gives A-2x 1 3; this, multiplied by the third term, 2 3 of a shilling, will give 2 3 of a shilling x 2- x 3. To reduce this with the least labor, we must resort to the method of canceling. Thus, canceling the 23, which occurs in both numerator and denominator, also 13 of the numerator against a part of the 26 of the denominator, our expression will, by this means, become I of a shilling x 9 x -- = of a shilling. NOTE.-This method of canceling should be used when the nature of the question will admit, since it will always very much simplify the operation. 4. If 34 pounds of coffee cost 23 shillings, how much will 104 pounds cost? 10* 114 HIGHER ARITHMETIC. In this example, 2- =- shillings must be our third term, and, since 10 - -' — pounds must cost more than 31 —7 pounds, we must divide f-L- by 4, for the ratio, making it -6- x 2; this, multiplied by the third term, - shillings, will give 7 shillings x -%L- x 42; this becomes, after canceling, I of a shilling x ----- shillings 64 shillings. It may happen that the three given quantities are all of one denomination; still it will be found that two of them are of one kind, and the third one of another kind, which is like the answer. As an example: What tax must an income of $6000 pay, provided $100 pays $1'03? Here the three given quantities are $6000, $100, $1'03, which are all of the same species, viz., money. Nevertheless, the first and second are incomes, and the third is a tax. So that the ratio must be $6000+-$100 -- 6L~-o —-= 60; therefore the tax sought is 60 times $1'03 -$61'80. 5. If a tree 38 feet 9 inches in height give a shadow of 49 feet 2 inches, how high is that tree, which, at the same time, casts a shadow of 71 feet 7 inches? In this example, our third term is the height of the first tree, which is 38 feet 9 inches=38 3 feet — l4- feet. Our ratio will be obtained by dividing 71 feet 7 inches= 71 -72 feet =- 8-5-? feet, by 49 feet 2 inches = 49- feet = 2 65 feet, which becomes 8529 -x; this, multiplied by the third term, 4]5 feet, gives 15 feet x 9 X5 Canceling 6 of the numerator against a part of the 12 of the denominator; also canceling 5, a factor of 155 of the numerator, against 5, a factor of 295 of the denominator, we get -3 feet x-5 x - -- 2 6 - =56 7 feet, 4 for 5the answ47 47er. for the answer. RULE OF THREE. 115 6. If 1617 yards of calico cost 213 shillings, how much can be bought for 32` shillings? Ans. 24 369 yards. 7. Sold a tankard for ~5 6s. at the rate of 5 s. 4d. per ounce. What was its weight? Ans. 1 lb. 7 oz. 17 pwt. 12 gr. -8. If 300 men consume 70 barrels of provisions in 10 months, how much will 240 men consume in the same time? Ans. 56 barrels. 9. Gave 72 dollars for 5 barrels of fish. How much will 20 barrels cost at the same rate? Ans. $288. 10. If it take 30 yards of carpeting, which is 3 of a yard wide, to cover a floor, how many yards, which is 1 I yards wide, will it take to cover the same floor?. Ans. 18 yards. 11. If an individual receive a salary of $700 per year, how much will that be for each day, counting 365 days for a year? Ans $1. 12. If the circumference of a wheel is 17- feet, what distance will it pass over in revolving 133 times? Ans. 233- feet. 13. If it take 2- yards of cloth for 2 pair of pantaloons, and 123 yards for 5 coats, how many yards will it require for 7 coats and 8 pair of pantaloons? Ans. 27A-4 yards. 14. If 19- yards of cloth cost 131 dollars, how many dollars will 14- yards cost? Ans. $95 4. 15. If 3 bushels of apples cost 13 shillings, how much will 172 bushels cost? Ans. ~3 15 s. 10 d. 16. How much must be paid for 33 cords of wood, at 22 dollars per cord? Ans. $8 — =$8'4375. 17. If a board 16 feet long and 9 inches wide contain 116 HIGHER ARITHMETIC. 12 square feet, how long must another board be, which is 8 inches wide, to contain the same number of square feet? Ans. 18 feet. 18. If the freight of 7 hogsheads of molasses for 18 miles is 9 dollars, what must be paid for the freight of 21 hogsheads the same distance? Ans. $27. 19. If 3 of a vessel is worth $1729, how much is the whole vessel worth? - Ans. $46103. 20. Lent my friend $300 for six months; afterwards he lent me $400. How long may I keep it to balance the favor? Ans. 4- months. 21. If a piece of land 20 rods long, and 8 rods wide, make one acre, how wide must it be, when it is 50 rods long, to contain the same? Ans. 3- rods. 22. If a ship sail 903 miles in 7- hours, how many hours will it require to sail 60 miles? Ans. 4 2 3 6 hours. 23. If 4 of an acre of land is worth 54 dollars, what is T- of an acre worth? Ans. $7'20. 24. If I pay 4 of a dollar for sawing one cord of wood, how much must I pay for sawing 53 cords? Ans. $4.125. If 16, yards of cloth are worth 2- of 7 of 204 dollars, what is that per yard? Ans. $2. 7 26. A man is $1700 in debt, and his estate amounts to but $870. How much can he pay on a dollar? Ans. $0'51 -327. How many yards of paper, 4 wide, will paper 375 square yards? Ans. 500 yards. 28. If a staff 5 feet long cast a shade 64 feet, what is the height of that steeple, whose shade at the same time measures 150 feet? Ans. 115-5A feet. RULE OF THREE. 117 29. If 3 paces, or common steps of a person, be equal to 2 yards, how many yards are there in 170 paces? Ais. 113- yards. 30. What cost 3 cwt. of coffee at 15 d. per pound? Ans. ~21. 31. A garrison of 540 men have provisions for 365 days. How long will those provisions last, if the garrison be increased to 1127 men? Ans. 174aoo days. 32. What will be the tax upon ~763 13 s. at the rate of 3 s. 4 d. on one pound sterling? Ans. ~1275s. 6d. 33. What is the value of a year's rent of 547 acres of land, at the rate of 15 s. 6 d. per acre? Ans. ~423 18 s. 6 d. 34. Allowing the French metre to be 3-75 feet in length, how many feet are there in 46 metres? Ans. 3-27 feet x 4-6 - - 2feet x 46 = 1502 2 feet. 35. Suppose a certain quantity of hay will feed 70 sheep 31 days, how long would it keep 131 sheep? Ans. 161-7-T days. 36. If 9 yards of silk, which is 3 of a yard wide, line a cloak, how many yards that is 4 wide, will it take to line the same? Ans. 9 yards x 3 x = —9 yards x - =52 yards. 37. If a barrel of beer last 10 men 16 days, how long will it last 27 men? Ans. 52 5 days. 38. If 92 barrels of flour are consumed by a company in 18 days, how long will 353 barrels last? 351! Ans. 18 days x =18 days x 3x- =69 30 days. 92 4 39. If a mill grind 19- bushels of corn in 1 hour, 17 minutes, in what time will it grind 100 bushels? 118 HIGHER ARITHMETIC. 19 =-7-7-; 1 hour, 17 minutes 77 minutes. Hence, we have, Ans. 77 mirn, x!Io0 x -7 = 400 min. = 6 h. 40 min. 40. If a barrel of flour will support 13 men for 27 days, how long would it support 9 men? Ans. 39 days. 41. If A of an acre of land cost $13, how much can be bought for $39? Ans. 1 7 acres. 42. If — A of a dollar will pay for 3 of a bushel of apples, how many bushels can be bought for 77- dollars? Ans. 312 bushels. 43. If 750 barrels of cider cost $2250, how much will 419 barrels cost? Ans. $1257. 44. If 1-3 of a firkin of butter is worth $1'80, what is of a firkin worth? Ans. $3'855. 45. If a staff 3 feet in length give a shadow 7 feet long, how high is that tree, which, at the same time, casts a shadow of 90 feet? Ans. 384 feet. 46. A regiment of soldiers, consisting of 976 men, is to be clothed, each coat to contain 2} yards of cloth, 1 5 yards wide, and to be lined with shalloon 7 of a yard wide. How many yards of shalloon will be required? Since each coat is to contain 2- = - yards, the number of yards for the whole regiment will be 976 x 2. Our expression will be 5 yards x 9 7 _ x 15 x =- yards x 1 6 X 1 — =4531 3 yards. 47. A person owning 3 of a coal-mine, sells ] of his share for $400. WThat is the whole mine worth at the same rate? Ans. $8881. 48. A and B hire a pasture for $50, in which A pastures 13 cows, and B 12. What must each pay? The whole number of cows pastured is 13+12=25. COMPOUND PROPORTION. 119 The ratio of A's to the whole is 3. The ratio of B's to the whole is, 2. Hence, A must pay $50 x -3 $26. B must pay $50 x - =$24. 49. Suppose sound to move 1106 feet in a second; how many miles distant is a cloud in which lightning is observed 47, seconds before the thunder is heard, no allowance being made for the progressive motion of light? Ans. 9 o 3 miles. 50. If A can mow an acre of grass in 52 hours, and B can mow 13 acres in 8I hours, in what time can they jointly mow 83 acres? Since A can mow 1 acre in 52 = hours, he can mow -7 =-51T o an acre in 1 hour. And, since B can mow 1 = 4 acres in 8' —'-7- hours, he can mow 4. — =81 of an acre in 1 hour. A and B can, together, mow -At +~ - _ — of an acre in 1 hour. Hence, to mow 83 =-3-5 acres, they will require 4 = 35 X 3 = 1 0 5 = 261 hours. COMPOUND PROPORTION. 62. WHEN the quantity required depends upon more than three terms, the operation of finding it is called the rule of compound proportion. Suppose we have the following example: If 20 men, working 10 hours each day, have been employed 18 days in constructing 500 feet of railroad, how -many days, of 12 hours each, must 76 men be employed to construct 1140 feet of the same road? Had the number of feet of road, as well as the num 120 HIGHER ARITHMETIC. ber of hours each day employed in labor, been the same in both cases, the question would have been equivalent to the following: If 20 men have been employed 18 days to perform a certain work, how many days would 76 men require to accomplish the same work? It is evident that the time sought in this case is the same fractional part of 18 days that 20 men is of 76 men; that is, the time required is 20 of 18 days. If, now, we take into account the number of hours employed each day, still supposing the number of feet of road to remain the same in both cases, our question will read thus: If it require 42 of 18 days to accomplish a certain work, when 10 hours are each day devoted to labor, how many days will be necessary when 12 hours are reckoned each day? The answer, in this case, is obviously Ao of 42 of 18 days. Now, taking into the account the number of feet of road, our question will become as follows: If I o of 24- of 18 days are required to construct 500 feet of railroad, how many days will be required to con struct 1140 feet? This leads to the following final result: J70- of I! of ] of 18 days-9 days. From the above work we see that questions of compound proportion may be solved by the following COMPOUND PROPORTION. 121 RULE. Among the terms given, there will be but one like the answer, which we will call the odd term. The other terms will appear in pairs, or couplets. Form ratios out of each couplet in the same manner as in the Rule of Three; then multiply the odd term and all the ratios together, and it will give the answer in the same name and denomination as the odd term. NoTE. —Before forming ratios from the couplets, they must be reduced to the same denominate value. EXAMPLES. 1. If a person travel 300 miles in 17 days, traveling only 6 hours each day, how many miles could he have gone in 15 days, by traveling 10 hours each day? In this example, the answer is required in miles; therefore, our odd term is 300 miles. The first couplet consists of days; and, since in 15 days, other things being the same, he could not travel as far as in 17 days, we must divide 15 by 17, which gives -5 for the first ratio. The second couplet consists of hours; and, since in 10 hours he could travel farther than in 6 hours, we must divide 10 by 6, which gives — ~- for the second ratio. Multiplying these two ratios and the odd term together, we get 300 miles x 1 5 x —. Canceling the 6 of the denominator against a part of 300 of the numerator, it becomes 50 miles x 4- 5x.A = 441 l- miles, for the answer. 2. If a marble slab, 10 feet long, 3 feet wide, and 3 inches thick, weigh 400 pounds, what will be the weight 11 122 IIIGHIER ARITHMETIC. of another slab, of the same marble, whose length is 8 feet, width 4 feet, and thickness 5 inches? In this example, the answer is required to be in pounds; therefore, 400 pounds is the odd term. The first couplet consists of the lengths; and, since 8 feet in length will give less weight than 10 feet, we must divide 8 by 10, which gives -8, for the first ratio. The second couplet consists of widths; and, since 4 feet wide will give more weight than 3 feet, we must divide 4 by 3, which gives 4 for the second ratio. The third couplet consists of thicknesses; and, since 5 inches thick will give more weight than 3 inches, we must divide 5 by 3, which gives 5 for the third ratio. Multiplying the odd term and these ratios together, we get 400 lbs. x - x - x 4. Canceling the 10 of the denominator against a part of the 400 of the numerator, we get 40 lbs. x I x 4 x 5 = — 49-00 = 711- pounds, for the answer. 3. If a pile of wood, 8 feet long, 4 feet wide, and 4 feet high, contain 1 cord, how many cords are there in a pile 26 feet long, 8 feet wide, and 12 feet high? In this example, 1 cord is the odd term. The first couplet consists of lengths; and, since 26 feet long will give more wood than 8 feet, we shall have _-96 for the first ratio. The second couplet consists of widths; and, since 8 feet wide will give more than 4 feet, we get 48 for the second ratio. The third couplet consists of heights; and, since 12 feet high will give more than 4 feet, we get J-Lg for the third ratio. Multiplying these ratios and the odd term, 1 cord, COMPOUND PROPORTION. 123 together, we get 1 cord x 298 x I x —. Canceling the 8 of the numerator against the 8 of the denominator, also one of the 4's of the denominator against a part of the 12 of the numerator, and the factor 2, of the remaining 4 of the denominator, against the factor 2 of 26, in the numerator, our expression, by this means, becomes, 1 cord x j-3 X 3=19.- cords, for the answer. 4. If 11 men can cut 49 cords of wood in 7 days, when they work 14 hours per day, how many men will it take to cut 140 cords in 28 days, by working 10 hours each day? Our expression will become, Ans. 11 men x'490 X X 4 1 1 men. 5. If 100 men, in 40 days, of 10 hours each, build a wall 30 feet long, 8 feet high, and 2 feet thick, how many men must be employed to build a wall 40 feet in length, 6 feet high, and 4 feet thick, in 20 days, by working 8 hours each day? o 0 0. - 0!1 o0 Ans. 100 men x x 1o X 4 o X X = 500 men. 6. If a man perform a journey of 1250 miles in 17 days, by traveling 13 hours a day, how many days will it take him to perform a journey of 1007 miles, by traveling 10 hours each day? Ans. 17' o o 04 days. 7. If 10 cows eat 8 tons of hay in 6 weeks, how many cows will eat 56 tons in 21 weeks? Ans. 20 cows. 8. If 8 men will mow 36 acres of grass in 9 days, 124 HIGHER ARITHMETIC. by working 9 hours each day, how many men will be required to mow 48 acres in 12 days, by working 12 hours each day? Ans. 6 men. 9. If 12 ounces of wool make 2- yards of cloth, which is 6 quarters wide, how many pounds will it take to make 150 yards of cloth, 4 quarters wide? Ans. 30 pounds. 10. If 12 men can build a wall 26 feet long, 7 feet high, and 5 feet thick, in 20 days, in how many days will 28 men build a wall 156 feet long, 10 feet high, and 3 feet thick? Ans. 44-49 days. 11. If the wages of 6 men for 14 days be $84, what will be the wages of 9 men for 16 days? Ans. $144. 12. If a pile of wood, 30- feet long, 4 feet wide, and 6 feet high, is worth $25, how much is a pile 60 feet in length, 3 feet wide, and 4 feet high, worth? Ans. $243 13. If 168 bushels of corn, when corn is worth 60 cents a bushels, be given for the carriage of 120 barrels of flour 60 miles, how many bushels of corn, when corn is worth 70 cents a bushel, must be given for the carriage of 80 barrels of flour 230 miles? Ans. 368 bushels. 14. A wall was to be built 700 yards long in 29 days; after 12 men had been employed on it for 11 days, it was found they had built only 220 yards. How many more men must be put on it to finish it in the given time? 700 yds. = whole length of wall. 220 yds. =part built in 11 days. 480 yds.=part to be built in the remaining 29-11 = 18 days. The question, therefore, may take this more COMPOUND PROPORTION. 125 simple form: If 12 mel in 11 days build 220 yards of wall, how many men will be necessary to build 480 yards in 18 days? This, when wrought by the rule, gives 12 41 —16 men. Now, as there are already 12 men upon the work, it becomes necessary to add 4 men more. 15. In how many days, working 9 hours a day, will 24 men dig a trench 420 yards long, 5 yards wide, and 3 yards deep, if 248 men, working 11 hours a day, in 5 days, dig a trench 230 yards long, 3 yards wide, and 2 yards deep? Ans. 5days1 5 x =2Sjr% days. 16. If 25 pears can be bought for 10 lemons, and 28 lemons for 18 pomegranates, and 1 pomegranate for 48 almonds, and 50 almonds for 70 chestnuts, and 108 chestnuts for 2~ cents, how many pears can I buy for $1'35? An 228 50 l0 - 3 135_7 Ans. 25 pears x x -I X X -7 o X 375 pears. 21 17. Suppose that 50 men, by working 5 hours each day, can dig, in 54 days, 24 cellars, which are each 36 feet long, 21 feet wide, and 10 feet deep, how many men would be required to dig, in 27 days, 18 cellars, which are each 48 feet long, 28 feet wide, and 9 feet deep, provided they work only 3 hours each day? Ans. 200 men. 18. If 15 men eat 13 shillings' worth of bread in 6 days, when wheat is sold at 12 shillings per bushel, in how many days will 30 men eat 43- shillings' worth, when wheat is 10 shillings per bushel? 43' Ans. 6 days x 3 x 5 - x 2=- 12 days. 1 13 126 HIGHER ARITHIMETIC. CHAPTER VI. SIMPLE INTEREST. 63. INTEREST is money paid by the borrower to the lender for the use of the money borrowed. It is estimated at a certain per cent. per annum; that is, a certain number of dollars for the use of $100, for one year. Thus, when $6 is paid for the use of $100, for one year, the interest is said to be at 6 per cent. In the same manner, when $5 is paid for the use of $100, for one year, the interest is said to be at 5 per cent.; and the same for other rates. The rate per cent. is generally fixed by law. In the New England States, the legal rate is 6 per cent., whilst in the State of New York, it is 7 per cent. The- sum of money borrowed, or the sum upon which the interest is computed, is called the principal. The principal, with the interest added to it, is called the amount. Case I. To find the interest on $1 for any given time at 6 per cent. The interest on $100, for one year, at 6 per cent., being $6, it follows that the interest on $1, for 1 year, is l~ of $6=$0'06; and, since 2 months is - -6 of one SIMPLE INTEREST. 127 year, the interest on $1 for 2 months, is 6 ot $0O06$0'01. Again, since 6 days is -= l of 2 months, when we reckon 30 days to each month, it follows that the interest on $1, for 6 days, is l-l of $0'01 =-$0001. Hence, we have the following RULE. Call half the number of months, CENTS; one-sixth the number of days, MILLS. EXAMPLE S. 1. What is the interest of $1 for 7 months and 10 days, at 6 per cent.? In this example, half the number of months is 3~; which, being called cents, gives $0'035 for the interest of $1 for 7 months; again, one-sixth of the number of days is 1U, which, being called mills, gives $0'0012- for the interest of $1 for 10 days; therefore, the interest for $1, for 7 months and 10 days, is $0'035+$0'001-= $0'036]3. 2. What is the interest of $1 for 11 months and 11 days, at 6 per cent.? Ans. $0-0565. 3. What is the interest of $1 for 3 years, 7 months, that is, for 43 months, at 6 per cent.? Ans. $0'215. 4. What is the interest of $1 for 2 years, 7 months, and 9 days, at 6 per cent.? Ans. $0 1565. 5. lihat is the interest of $1 for 1 year, 7 months, and 15 days, at 6 per cent.? Ans. $0'0975. 6. What is the interest of $1 for 7 years and 9 days, at 6 per cent.? Ans. $0'4215. 128 HIGHER ARITHMETIC' Case II. To find the interest of any given principal, for any given time, at 6 per cent., we have this RULE. Find the interest on $1 for the given time, by Case L, Art. 63. Then multiply the interest thus found by the number of dollars in the given principal. EXAMPLES. 1. What is the interest of $49'37 for 13 months and 15 days, at 6 per cent.? In this example, we find the interest on $1'for 13 months and 15 days, at 6 per cent., to be $0'0675; which, multiplied by $49'37, gives $3'332475, for the interest on $49'37, for the given time. 2. What is the interest of $608-62 for I year and 9 months, at 6 per cent.? Ans. $63'9051. 3. What is the interest of $341'13 for 7 years and 9 days, at 6 per cent.? Ans. $143'786295. 4. What is the interest of $100 for 16 years and 8 months, at 6 per cent.? Ans. $100. 5. What is the interest of $591i03 for 4 years, 3 months, and 7 days, at 6 per cent.? Ans. $151'402185. 6. What is the interest of $0'134 for 4 months and 3 days, at 6 per cent.? Ans. $0'002747. SIMPLE INTEREST. 129 Case III. To find the interest on any given principal, for any given time, at any given rate per cent., we have this RULE. Find the interest on the given principal, for the given time, at 6 per cent., by Case II., Art. 63. Then increase, or decrease, this interest by the same part of itself as it would be necessary to increase, or decrease 6, in order to make it agree with the given rate per cent. EXAMPLES. 1. What is the interest of $19'41 for 1 year, 7 months, ~and 13 days, at 7 per cent.? In this example, we find, by Case II., that the interest of $19'41 for 1 year, 7 months, and 13 days, at 6 ptr cent., is $1'886005. Since 6, increased by its sixth part, equals 7, it will be necessary to increase the interest just found, for 6 per cent., by its sixth part, which thus becomes $2'2003391, for the interest at 7 per cent. 2. What is the interest of $530 for 3 years and 6 months, at 5 per cent.? Ans. $92'75. In this example, it was necessary to decrease the interest at 6 per cent. by its sixth part. 3. What is the interest of $5'37 for 4 years and 12 days, at 8 per cent.? Ans. $1'73272. In this example, we increased the interest at 6 per cent. by its third part. 130 HIGHER AI l'l IE'TJ c. 4. What is the interest of $4070 for 3 months, at 9 per cent.? Ans. $91'575. 5. What is the interest of $3671 for 6 months, at 10 per cent.? Ans. $183'55. 6. What is the interest of $4920'05 for 3 months, at 4 per cent.? Ans. $49'2005. 7. What is the interest of $40'17 for 3 months and 18 days, at 3 per cent.? Ans. $0'36153. 8. What is the interest of $37'13 for 5 months and 12 days, at 4- per cent.? Ans. $0'7518825. NOTE. —When the principal is given in English money, we must reduce the shillings, pence, and farthings to the decimal of a pound, and then proceed as in Federal money. 9. What is the interest of ~75 13 s. 6 d. for 3 years and 5 months, at 6 per cent.? In this example, 13 s. 6 d., reduced to the decimal of a pound, is ~0'675, so that our principal is ~75'675; the interest on ~1 for 3 years and 5 -months, at 6 per dent., is ~0'205, which, multiplied into 75'675, the number of pounds, gives ~15'513375= —~15 10 s. 3_,4- d for the interest required. 10. What is the interest of ~14 5 s. 3- d. for 4 Vears, 6 months, and 14 days, at 7 per cent.? Ans. ~4 10 s. 7-c 1L., nearly. 11. What is the interest of ~1 7 s. 6 d. for 2 years and 6 months, at 4- per cent.? Ans. ~0 3 s. 1 d. 12. What is the interest of ~105 10 s. 6 d. for 91 months, at 5 per cent.? Ans. ~4 3 s. 6 d. 1 -95 far. When the interest is 7 per cent., it may be readily computed by the assistance of the following table, which gives the interest of $1, or ~1, for any time, in months and days, not exceeding 1 year. SIMPLE INTEREST. 131 INTEREST TABLE AT SEVEN PER CENT. Days. 0 Month. I Month, 2 Months. 3 Months. 4 Months. 5 Months. 0 0 000010K005830 01167 0 01750 002333 002917 1 o000190'00603 0 01 186 0o01 69 002353 002936 2 0'00039 0'00622 0'01206 0'01789 0'02372 0'02956 3 0'00058 0'00642 0'01225 0'01808 0'02392 0'02975 4'07007S 0-00661 0'01244 10'01828 0'02411 0 02994 5 0'00097 0'00681 0-01264 0 01847 0'02431 1003014 6 000117 0"00700 0'0128 0 01867 10'02450 0'03033 7 0'00136 0'00719 0'01 303 0'01886 1002469 0'03053 8 0'00156 0'00739 0'01322 0'01906 0'02489 0-03072 9 0'00175 0 00758 0'01342 10'01925 0'02508 O003092 10 0-00194 0'00778 0 01361 0 01944 0'02528 0303111 11 0'00214 0'00797 001381 0 01964 0-02547 0-03131 12 0O00233'00817 0'01400 0'01983 1002567 0[03150 113 0'00253 0'00836 0'01419 0'2003 002586 0-03169 14 0'00272 1000856 0'01439 /002022 0'02606 1003189 15 0'00292 0/00875 001458 002042 0'02625 1003208 16 0'00311 0'00894 0'01478 0'02061 0 02644 0 03228 17 0'00331 0'00914 0'01497 0'02081 0'02664 10'03247 18 0'00350 0OG933 0'01517 0O02100 0'02683 10'03267 19 0 00369 0'00953 0'01536 0'02119 0-02703 0'03286 20 0'00389 0'00972 0'01556 0'02139 0'02722 0'03306 21 000408 0'00992 0-01575 0'02158 O'02742 0'03325 22 0'00428 001011 0g01594 10'02178 0'02761 0'03344 23 00044710'01031 0'01614 0'02197 0'02781 0'03364 24 0-00467 0'01050 0-01633 0'02217 0'02800 0'03383 25 0'00486 0'01069 0-'01653 0-02236 0'02819 0'03403 26 0'00506 0'01089 0'01672 0'02256 0-02839 0'03422 27 0'00525 0'01 108 0'01692 0-02275 0'0285810'03442 28 0'005440-'01128 0-01711 0O02294 0-02878 0'03461 29 0'0056410'0114710'0173110-02314 0-02897 0-03481 30 0'00583 0-01167j0'0175010'02333 0-0291710-03500 132 IlTrICt;: ER ARITHM.IE';i( INTEREST TABLE AT SEVEN PER CENT. Days. 6 Months. 7Months. Months. 9 Momths, 10 Months. ll Months. o0 0'03500 004083 004667 005250 0-05833 0 06417 1 100351910'04103 0-04686 10052690'05853 0-06436 2 0'03&39 0-04122 0-04706 0-05289 0-05872 0-06456 3 0'03558 0-04142 0-04725F 0-053080-05s92 0-06475 4 003578 0-04161 0-04744 0-05228S0O05911 0-06494 5 0[03597 0-04181 1004764 0-05347 0-05931 0-06514 6 0-03617 0-04200 0-04783 0-05367 0'05950 0-06533 7 /003636 0'04219 0-0480310-05386 0'05969 0'06553 8 0-03656 0-04239 0-048'22 005406&0-05989' 006572 9 0-03675 0-04258 0-04842 0-05425 0-06008 0-06592 10 0-03694 0-04278 0'04861 0-05444 0'0602b80-06611 11 0-03714 0-04297 0-04881 0-05464 0-0604710-06631 12 0-03733 0'04317 0-04900 0-05483 0-06067 0-06650 13 0-03753 0-04336 0-04919 0-0550310-06086 006669 14 0-03772 0-04356 0-04939 0-05522 0-06106 0.06689 15 0-03792 0-04375 0-04958 0-05542 0o06125 0-06708 16 0-03811 0-04394 0-04978 0-05561 1006144 0-06728 17 0-03831 0-04414 0-04997 0-05581 0-06164'0-06747 18 0-03850 0-04433 0-05017 0-05600 0-06183 0-06767 19 0-03869 0-04453 0-05036 0-05619 0-06203 0-06786 20 0-0388910-04472 0-05056 0-05639 0-06222,0'06806 21 0-039051 0-04492 0-05075 0-05658 0-06242 0-06825 22 0-03928 0-04511 0-05094 0-05678 006261 0-06844 23 0-03947 0-04531 0-05114 0-05697 0-06281 0-06864 24 0-03967 0-04550 0-05133 0-05717 /0-06300 0-06883 25 0-03986 0-04569 0-05153 0-05736 00'06319 006903 26 0-04006 0-04589 0-05172 0-05756 0'06339 0-06922 27 0-0402510-04608l0-05192 0-05775 0-06358 0-06942 28 0-04044 0-04628 /005211 0-05794 0-06378 1006961 29 0-040641004647 0-05231!005814.0-06397 0-06981 30 0-04083 0-04667 9-0525010-05833 0-0641710-07000i SIM1PLEV IN7tEREST. 133 We will now give, to be wrought by the aid of the preceding table, the following EXAMPLE S. 1. What is the interest of $37'13 for 3 months and 3 days, at 7 per cent.? Operation. $0 01808 - - Tabular number, 3 mo. 3 days. 37'13 - - Multiply by the number of dollars 5424 in the principal. 1808 12656 5424 $0'6713104, Interest sought. 2. What is the interest of $320 for 6 months and 3 days, at 7 per cent.? Ans. $11'386, nearly. 3. What is the interest of ~20 5 s. 6 d. for 8 months 17 days, at 7 per cent.? Reducing the shillings and pence to the decimal of a pound, our principal will become ~20'275.. Hence, multiplying ~0'04997, the tabular number for 8 months 17 days, by 20'275, we find ~1'01314175. Reducing this decimal of a pound to shillings and pence, we have ~1 0 s. 3 d., nearly for the interest required. 4. What is the interest of $500 for 6 months and 1 day, at 7 per cent.? Ans. $17'595. 5. What is the interest of ~500 10 s. for 4 months 15 days, at 7 per cent.? Ans. ~13 2 s. 9T1- d. 6. What is the interest of $1250 for 3 years and 3 months, at 7 per cent.? 12 18411 11 I G -El1G E 1 ARI THIMETIC. The tabular number for 3 months is $0'01750; to which, adding $0'21, the interest of $1 for 3 years, we have $0'2275, for the interest of $1 for 3 years and 3 months. Now, multiplying $0'2275 by 1250, we obtain $284'375, for the interest sought. 7. What is the interest of $33'33 for 2 years, 5 months, and 3 days, at 7, per cent.? Ans. $5'658, nearly. PARTIAL PAYMENTS. 64. WVIEN notes, bonds, or obligations, receive partial payments, or indorsements, we must use the following Rule, which was given by CHANCELLOR KENT, in the New York Chancery Reports: RULE. " The Rule for casting interest, when partial payments have been made, is to apply the payment in the first place to the discharge of the interest then due. If the payment exceed the interest, the surplus goes towards discharging the principal, and the subsequent interest is to be computed on the balance of the principal remaining due. If the payment be less than the interest, the surplus of interest must not be taken to augment the principal; but interest continues on the former principal until the period when the payments, taken together, equal or exceed the interest due, and then the surplus is to be applied towards discharging the principal, and interest is to be computed on the balance, as aforesaid." PARTIAL PAYMENTS. 135 UTICA, May 16, 1839. 1. For value received, I promise to pay A. B., or order, three hundred and twenty dollars, with interest at 7 per cent. C. D. Indorsements were made on this note as follows: September 3, 1839, there was paid... $3000 January 5, 1840, " " "... 5'00 August 11, 1841, " " "... 4000 February 25, 1842, " " "... 100'00 How much was due May 3, 1843? We first find the successive periods of time for which interest is to be computed by the following Operation. Year. Mo. Day. Date of note, 1839 4 16 Mo. Day. 1st indorsement, 1839 8 3 3 17 2d indorselnent, 1840 0 5 Year. 4 2 1 7 6 3d indorsement, 1841 7 11 6 14 4th indorsement, 1842 1 25 1 2 8 Date of settlement, 1843 4 3 The interest being 7 per cent., we will compute it by the aid of the table under Art. 63. Amount of the note, or the principal, is ~ ~ $320'000 Interest up to Sept. 3, 1839, is... 6'659 326'659 Deduct first indorsement,....... 30'000 296'659 Interest up to January 5, 1840, is $7'037, which is greater than the second indorsement. 136 IIIGHER ARITHMETIC. Brought forward,........ $296-659 Interest from Sept. 3, 1839, to August 11, 1841, is....... 40'263 336'922 Sum of second and third indorsements,. ~ ~ 45000 291'922 Interest up to February 25, 1842, is. 11'011 302'933 Deduct fourth indorsement,.... 100'000 202'933 Interest up to May 3, 1843, is. ~. 16'888 Amount due May 3, 1843, is.. $219'821 UTICA, Nov. 1, 1837. 2. For value received, I promise to pay THOMAS JONES, or order, the sum of six hundred and twenty dollars, on demand, with interest. CHARLES BANK. The following indorsements were made on this note: 1838, Oct. 6, there was indorsed. ~. $61'07 1839, March 4, " " "... 89'03 1839; Dec. 11, " " ".. 107'77 1840, July 20, " " "...200'50 What was the balance due Oct. 15, 1840, allowing 7 per cent. interest? Year. Mo. Day. Date of note, 1837 10 1 Mo. Day. 1st indorsement, 1838 9 6 11 5 2d indorsement, 1839 2 4 4 28 9 7 3d indorsement, 1839 11 11 7 9 4th indorsement, 1840 6 20 2 25 Date of settlement, 1840 9 15 PARTIAL PAYMENTS. 137 The amount of note, or principal, is.. $620'000 Interest on the same, up to Oct. 6, 1838, is 40'386 Amount due on note, Oct. 6, 1838, is. 660'386 The first indorsement is.. 61'070 599'316 Interest from Oct. 6, 1838, to March 4, 1839, is 17'247 Amount due March 4, 1839, is. 616'563 The second indorsement is.... 89030 527'533 Interest from March 4, 1839, to Dec. 11, 1839, is 28'414 555.947 The third indorsement is..... 107770 448'177 Interest from Dec. 11, 1839, to July 20, 1840, is 19'085 467'262 The fourth indorsement is.. 200500 266'762 Interest from July 20, 1840, to Oct. 15, 1840, is 4'409 Ans $271'171 UTICA, May 1, 1836. 3. For value received, I promise to pay ISAAC CLARK, or order, three hundred and forty-nine dollars, ninetynine cents, and eight mills, with interest at 6 per cent. N. BROWN. Indorsements were made on this note as follows: Dec. 25, 1836, there was paid ~... $49'998 June 30, 1837, " " "..... 4998 Aug. 22, 1838 " " ".... 15'000 June 4, 1839, " " ".... 99999 12* 138 HIGHER ARITHMETIC. How much was due April 5, 1840? Year. 3M-o. Day. Interest at Interest at Date of note 1836 4 1 Mo. Day. 6 per cent. 1st indorsement, 1836 11 25 7 24 $0 039 2d (" 837 5 30 Year. 6 5 0'0301 1 22 0'068-2 3d " 1838 7 22 93 9 12 0-047 4th " 1839 5 4 10 1 0'050Date of settlement, 1840 3 5 The amount of the note, or principal, is.. $349'998 Interest up to Dec. 25, 1836, is..... 13650 363'648 The first indorsement is... 49'998 313'650 Interest up to June 4, 1839, is... 45'950 359'600 Indorsement June 30, 1837, which $4'99S is less than the interest then due, Indorsement August 22, 183S, ~. 15'000 19'998 Thlis sum is still less than the interest now due. Indorsement June 4, 1839, ~ ~. 99'999 $119'997 This sum exceeds the interest now due. 239'603 Interest up to April 5, 1840, is..... 12020 Amount due April 5, 1840,..... 251623 UTICA, Dec. 9. 18.35. 4. For value received, I promise to pay PETER SMITII, or order, one hundred and eight dollars and forty-three cents, on demand, with interest at 7 per cent. JOHN SAVEALL. PARTIAL PAYMENTS. 139 Indorsements were made as follows: March 3, 1836, there was indorsed.. $50'04 Dec. 10, 1836, " " ". 13'19 May 1, 1838, " " ". 50 11 How much remained due Oct. 9, 1840? Ans. $5s844. UTICA, Aug. 1, 1837. 5. For value received, I promise to pay F. GOULD, or bearer, one hundred and forty-three dollars and fifty cents, on deinani, with interest D. FARLING. Dec. 17, 1837, there was indorsed. $37'40 July 1, 1838, " " ".... 709 Dec. 22, 1839,' " " 1313 Sept. 9, 1840, " " " ~. 50'50 How much remains due Dec. 28, 1840, the interest being 7 per cent.? Ans. $60'866. 6. A note of $486 is dated Sept. 7, 1831, on which, March 22, 1832, there was paid.... $125 Nov. 29, 1832, " " ".. 150 May 13, 1833, " " "..... 120 What was the balance due April 19, 1834, the interest being 7 per cent.? Ans. $144'404. 65. The principal, the rate per cent., the time, and the interest, are so related to each other, that any three of them being given, the remaining one can be found. 140 HIGHER ARITHMETIC. Problem I. Given the principal, the rate per cent., and the time, to find the interest. The rule for this problem has already been given under Case II!., Art. 63; it is equivalent to the following RULE. Multiply the interest of $1 for the given time and given rate per cent., by the numberf dollars in the principal. Problem II. Given the time, the rate per cent., and the interest, to find the principal. By the reverse of the last problem,. we obtain this RULE. Divide the given interest by the interest of $1 for the given time and given rate per cent.; and the quotient will be the number of dollars in the principal. EXAMPLES. 1. The interest on a certain principal for 9 months and 10 days, at 41 per cent., is $1'01605. What was the principal? In this example, we find the interest of $1 for 9 months and 10 days, at 4- per cent., to be $0035;.'. $1'01605, divided by $0'035, gives 29'03 for the number of dollars in the principal required. 2. What principal will, in I year, 7 months, and 15 days, at 6 per cent., give $9'75 interest? Ans. $100. PROBLEMS IN INTEREST. 141 3. What principal will, in 7 years and 9 days, at 6 per cent., give $16'86 interest? Ans. $40. 4. What principal will, in 3 years and 6 months, at 5 per cent., give $92'75 interest? Ans. $530. 5. What principal will, in 3 months and 9 days, at 8 per cent., give $90, interest? Ans. $4090'909. Problem III. Given the principal, the time, and the interest, to find the rate per cent. RULE. Divide the given interest by the interest of the given principal, for the given time, at one per cent. EXAMPLES. 1. The interest of $100 for 9 months and 10 days, is $3'50. What is the rate per cent.? In this example, we find the interest of $100 for 9 months and 10 days, at 6 per cent., to be $4'662-. The interest at 1 per cent. is 6 of $4'66-3=$0'777; therefore, dividing $3'50 by $0 777, we obtain 41 for the rate per cent. required. 2. At what rate per cent. will $530, in 3 years and 6 months, give $92'75, interest? Ans. 5 per cent. 3. At what rate per cent. will $19'41, in 1 year, 7 months, and 13 days, give $2 2003391, interest? Ans. 7 per cent. 4. At what rate per cent. will $5'37, in 4 years and 12 days, give $1 73272, interest? Ans. 8 per cent. 142 HIGHER ARITHMETIC. 5. At what rate per cent. will $4070, in 3 months, give $91'575, interest? Ans. 9 per cent. Problem IV. Given the principal, the rate per cent., and the interest, to find the time. RULE. Divide the given interest by the interest of the given principal for 1 year at the given rate per cent. EXAMPLES. 1. In what time will $37'13, at 44 per cent., give $0'7518825, interest? In this example, we find the interest of $37'13 for 1 year, at 4- per cent., to be $1'67085;.'. dividing $0"7518825 by $1'67085, we get 0'45, which, considered as years, gives, when reduced, 5 months and 12 days. 2. In what time will $700, at 7 per cent., give $85'75, interest? Ans. 1 year, 9 months. 3. In what time will $100, at 6 per cent., give $100 interest? That is, in what time will a given principal double itself at 6 per cent. interest? Ans. 162 years. 4. In what time will a given principal double itself at 7 per cent.? Ans. 1424 years. -5. In what time will a given principal double itself at 8 per cent.? Ans. 12- years. 6. In what time will a given principal double itself at 5 per cent.? Ans. 20 years. 7. In what time will a given principal double itself at 44- per cent.? Ans. 222 years. DISCOUNT. 143 To find the number of years required for a given sum to double itself at simple interest, we have only to divide 100 by the rate per cent. The following table gives the time required for a given principal to double itself at simple interest. Per cent. Years. Per cent. Years. Per cent. Years. 1 100 4 25 7 142 1'{ 662 4- 222 71 132 50 5 20 8 1221 40 5- 18, 82 11 - 242 1 21 1 T 331 6 162 9 1131 284 6- 15jk- 9 10'~ DISCOUNT. 66. DISCOUNT is an allowance made for the payment of money before it is due. The present worth of a debt payable at some future time, without interest, is such a sum of money as will, in the given time, amount to the debt. When the interest is at 6 per cent., the amount of $1 for one year, is $1-06; therefore, the present worth of $1'06, due one year hence, is $1. We may also infer that the present worth of any sum for one year will be as many dollars as $1'06 is contained in the given sum. Hence, we have the following RULE. Find the amount of $1 for the given time, at the given rate per cent.; then divide the sum by this 144 HIGHER ARITHMETIC. amount, and it will give the number of dollars in the present worth. Subtract the present worth from the amount, and it will give the discount. EXAMPLES. 1. What is the present worth of $622'75, due 3 years and 6 monttis, at 5 per cent.? In this example, we find the amount of $1 for 3 years and 6 months, at 5 per cent., to be $1'175; therefore, dividing $622'75 by $1'175, we get 530 for the number of dollars in the present worth. If we subtract the present worth from the sum, we get $92'75 for the discount. 2. What is the present worth of $4161'575, due 3 months hence, at 9 per cent.? Ans. $4070. 3. What is the present worth of $7'10272, due 4 years and 12 days hence, at 8 per cent.? Ans. $5'37. 4. Sold goods for $1500, to be paid one half in 6 months, and the other half in 9 months; what is the present worth of the goods, interest being at 7 per cent.? Ans. $1437'227. 5. What is the present worth of $50, payable at the end of 3 months, at 7 per cent.? Ans. $49'14. 6. What is the discount on $100, due 6 months hence, at 6 per cent.? Ans. $2'91'3. The following table gives the present worth of $1, or ~1, for months and days, for any time not exceeding 1 year, at 7 per cent. DIJSCO UN T. 145 TABLE OF PRESENT WORTHS. Days. O Molnth. 1 Month. 2 Months. 3 Months. 0 0'000000 0-99420() 0'988468 0'982801 1 0'999806 0'994008 i. 0-988278 0'982613 2 0'999600 0'993816 0'988088 0'982425 3 0'999417 0'993624 O 0987898 0'982238 4 0'999223 0'993432 0'987709 0'982050 5 0'999029 0'99'3240 0-987519 0'981863 6 0'998835 0'993049 0'987329 0981675 7 0'998641 0992857 0987140 0981488 8 0'998447 0'992665 0'986950 0'981301 9 0'998253 0.992474 0986761 0'981 114 10 0'998059 0'992282 0O986572 0'980926 1 01 0'997866 0'992091 0'986382 0'980739 12 0'997672 ()991899 0'6193 0980552 13 0'997479 0'991708 0-986004 0'980365 14 0'997285 0'991517 09()95815 0980179 15 0'997092 0'991 326 ()'985626 0'979992 16 0'996899 0'9911;35 0'985437 1 0'979805 17 0'996705 0'990944 0()'3249 0(-979618 18 0'996512 0'990753 0'9850G60 0'979432 19 0'996319 00'99056 0 981471 0'979245 20 0'996126 0'990371 0'984683 i 0-97U059 21 0'995933 0'990181 0(984494 0'978873 22 0'995740 0'989990 0'984306 0'978686 23 0'995548 ()0989500 0()984117 0'97850.0 24 0'995355 0'989609 0'983t92 9 0'978314 25 0'995162 0'989419 0 983741 0-978128 26 0'994970 0'989228 0('983553 0977942 27 0'994777 0'989038 0'983365 0'977756 28 0'994585 0'988848 0'983177 0'977570 29 0'994393 0'988658 0'982989 0977384 30 0'994200 0'988468 0'982801 0977199 146 HIIGHIER ARITIMETIC,. TABLE OF PRESENT WORTHS. Days. 4 Months. 5 Months. 6 Months. 7 Months. 0 0'977199 0'971660 0'966184 0'960769 1 0'977013 0'971476 0'966002 0'960589 2 0'976827 0'971298 0'965821 0'960410 3 0'976642 0'971109 0'965639 -0'960230 4 0'976457 0'970926 0'965458 0'960051 5 0'976271 0'970743 0'965277 0'959872 6 0'976086 0'970560 0'965096 0'959693 7 0-975901 0'970377 0'964915 0'959514 8 0'975716 0'970193 0'964734 0'959335 9 0'975530 0'970011 0'964553 0-959156 10 0'975345 0'969828 0-964372 0-958977 11 0'975160 0'969645 0-964191 0'958798 12 0'974976 0'969462 0'964010 0'958620 13 0'974791 0'969279 0'963830 0'958441 14 0'974606 0'969097 0'963649 0'958262 15 0'974421 0'968914 0-963468 0'958084 16 0'974237 0'968731 0'963288 0'957905 17 0'974052 0'968549 0'963108 0'957727 18 0'973868 0'968367 0'962927 0'957549 19 0'973683 0'968184 0'962747 0'957370 20 0'973499 0'968002 0'962567 0'957192 21 0'973315 0-967820 0'962387 0'957014 22 0`973131 0'967638 0'962207 0'956836 23 0'972947 0'967456 0'962027 0'956658 24 0'972763 0'967274 0'961847 0'956480 25 0'972579 0'967092 0'961667 0'956302 26 0'972395 0'966910 0'961487 0'956125 27 0'972211 0'966728 0'961307 0'955947 28 0'972027 0'966547 0961128 0'955769 29 0'971S44 0'966365 0'960948 0'955592 30 0'971660 0'966184 0'960769 0'955414 I 097027 0~665W ~96 15 0~57i DISCOUNT. 147 TABLE OF PRESENT WORTHS. Days. 8 Months. 9 Months. 10 Months. 11 Months. 0 0'955414 0'950119 0'944882 0'939702 1 0'955237 0'949943 0'944708 0'939531 2 0'955059 0'949768 0'944535 0'939359 3 0'954882 0'949592 0'944361 0'939188 4 0'954705 0-949417 0'944188 0'939016 5 0'954527 0'949242 0'944015 0'938845 6 0'954350 0'949067 0'943841 0'93S673 7 0'954173 0'948892 0'943668 0'938502 8 0'953996 0'948717 0'943495 0'938331 9 0'953819 0'948542 0'943322 0'938160 10 0'953642 0'948367 0'943149 0'937989 11 0'953466 0'948192'0'942976 0937817 12 0'953289 0 948017 0'942803 0'937647 13 0'953112 0'947842 0'942630 0'937476 14 0'952936 0'947668 0'94245S 0'937305 15 0'952759 0'947493 0'942285 0'937134 16 0'952583 0'947319 0'942112 0'936963 17 0'952406 0'947144 0'941940 0'936793 18 0'952230 0'946970 0'941.767 0'936622 19 0'952054 0'946795 0'941595 0'936451 20 0'951877 0'946621 0'941423 0'936281 21 0'951701 0'946447 0'941250 0'936110 22 0'951525 0'946273 0'941078 0'935940 23 0'951 349 0'946099 0'940906 0'935770 24 0'951173 0'945925 0'940734 0'935600 25 0'950997 0'945751 0'940562 0'935429 26 0'950821 0'945577 0'940390 0'935259 27 0'950646 0'945403 0'940218 0'935089 28 0'950470 0'945229 0'940046 0'934919 29 0'950294 0'945056 0'939874 0'934749 30 0'950119 0'944882 0'939702 0'934579 148 111G HER ARITIHMETIC. The following examples may be worked by the aid of the foregoing table: 7. What is the present worth, interest being 7 per cent., of $320, due at the end of 6 months and 3 days? Operation. $0965639 - -- i Tabular number for 6 months 3 days. 320 - -- -Multiply by the number of dollars. 19312780 2896917 Ans. $309'004480 8. What is the discount on $750, due 9 months hence, at 7 per cent.? Ans. $37'411. 9. What is the present worth of $3471'20, due 3 months and 9 days hence, at 7 per cent.? Ans. $3405'643. 10. What is the discount of $150, due 3 months and 18 days hence, at 7 per cent.? Ans. $3'085. 11. What is the discount of $961'13, due 10 months and 5 days hence, at 7 per cent.? Ans. $53'809. EQUATION OF PAYMENTS, DEDUCED BY CONSIDERING THE PRESENT VALUES OF THE SEVERAL PAYMENTS. 67. SUPPOSE A owes me $100, due at the end of 3 months, and $100, due at the end of 9 months, and I wish him to give me one note of $200, of such a time that its present value shall be the same as the sum of the present values of the two individual debts. How long after date must this note be made payable? EQUATION OF PAYMENTS. 149 By the foregoing table for the present worth, we find the present value of $100 for three months, to be $98'2801: the present value of $100 for 9 months, to be $95'0119. Taking the sum, we have $193'292 for the present value of $200, due at a future time, which time we are required to find. We may obviously consider $193'292 as a principal, which, at the given rate per cent., will amount to $200, in the time sought. The interest is, therefore, $6-708. This question now is equivalent to the following: - Given the principal, the rate per cent., and the interest, to find the time. This has been solved under Prob. IV., Art. 65. Proceeding according to this rule, we find the interest of $193'292 for 1 year, at 7 per cent., to be $13'53, nearly. Dividing $6'708 by $13'53, we get 0-4958 of a year, equal to 5 months, 28S49 days, nearly. The equated time, when found by the ordinary method, is 6 months. From the above, we deduce the following rule for the equation of payments, founded on the principle of equivalent present values. RULE. Find the sum of the present values of the individual debts; also, the sum of their discounts. Then regard the sum of the present values of the individual debts as a principal, and the sum of their discounts as the interest. Then proceed with this principal and interest and given rate per cent., according to Rule under Prob. IV., Art. 65. 13* 150 HIGHlER ARLTIIME iIC. EXAMPLE. Suppose A owes me $500, due at the end of 3 months, $600 at the end of 4 months, and $800 at the end of 6 months. How long may the whole $1900 remain unpaid, so that its present worth may be the same as the sum of the present worths of the individual debts? Present value of $500, due 3 months hence, is $191'4005 " " 6(00, " 4 " " 586'3194 "' " c 800, " 6 " " 7729472 Sum of their present values is... $1850'6671 Sum of several payments $1900'0000 1850'6671 Sum of their discounts is $49'3329 Now, if a principal of $1850'6671 gives $49'3329 interest at 7 per cent., what is the number of years? By rule under Prob. IV., we find that the interest on $1850'6671 for 1 year, at 7 per cent., to be $129'5467. Hence, dividing $49'3329 by $129'5467, we find 0'3808 of a year, which is the same as 4 months, 17'088 days, nearly. If we find the equated time by the usual rule for equation of payments, it will be 4 months, 17'368 days, nearly, which differs less than half a day from the result by the above method. Other examples might be given to illustrate this method of finding the time for the equation of payments, on the principle of equivalent present values, but enough has been done to call the attention of the student to this singular subject. COMPOUND INTEREST. 151 CHAPTER VI I. COMPOUND INTEREST. OS. WHEN, at the end of each year, the interest due is added to the principal, and the amount thus obtained is considered as a new principal, upon which the interest is cast for another year, and added to it to form a new principal for the next year, and so on to the last year, the last amount thus obtained, is called the AMOUNT AT COMPOUND INTEREST. If from this amount we subtract the original principal, we obtain the COMPOUND INTEREST. EXAMPLES. 1. What is the compound interest of $1000 for 3 years, at 7 per cent.? Principal,.... $1000 Interest on $1000 for one year,... 70 First amount, or second principal,. ~ 1070 Interest on $1070 for one year,... 74'90 Second amount, or third principal,.. 1144'90 Interest on $1144'90 for one year,.. 80'143 Third amount,.... 1225'043 Original principal,...... 1000 The compound interest required, ~ Ans. *$225'043 -152 I1 ( 11'I It A R I' ri E'l M 1J C. 2. What is the compound interest of $100 for 4 years, at 6 per cent.? Principal,.... $100 Interest for first year.... 6 First amount, or second principal.... 106 Interest for second year,..... 636 Second amount, or third principal,.. 112'36 Interest for third year,..... 674 Third amount, or fourth principal,. 119-10 Interest for fourth year,... 7 15 Fourth amount,..126'25 Original principal,.1. 100 Compound interest required,.. Ans. $26'25 3. What is the compound interest of $630 for 4 years, at 5 per cent.? Ans. $135'769. By carefully reviewing the above manner in which compound interest is computed, we discover that the'successive amounts, which are considered as new principals, form the terms of a geometrical series, whose first term is the original principal; the ratio is the amount of $1 for one year, at the given rate per cent.; the number of terms is equal to the number of years, plus one. From this we learn that finding the amount of a given principal, for a given number of years, at a given rate per cent., consists in finding the last term of a geometrical progression, when the first term, the ratio, and the number of terms are given. Thus, the amount of $1 for one year, at 3 per cent., is $1'03; for two years, it is $(103)2; for three years, it is $(1'03) 3; for four years, it is $(1'03)4; and in general, for any number of years, it is found by raising 1'03 to a power denoted by the number of years. Co oIlOttUNT INTEItEST. 153 TAB LE, SHOWING the armount o;' $1, or ~1, for any number of years, not exceeding 30, at 3, 4, 5, and 6 per cent., at compound interest, the interest being compounded yearly. Years. 3 per cent. I per cent. 5 per cent. 6 per cent. 1 1'030000 1'040000 1'050000 1'060000 2 1'060900 1'081600 1'102500 1'123600 3 1'092727 1'124864 1'157625 1'191016 4 1'125509 1'169859 1'215506 1'262477 5 1'159274 1 216653 1'276282 1'338225 6 1'194052 1'265319 1'340096 1'418519 1'229874 1'315932 1 1'407100 1'503630 8 1'266770 1'368569 1'477455 1'593848 9 1'304773 1.423312 1'551328 1'689479 10 1'343916 1'480244 1'628895 1'790848 11 1'384234 1'539454 1'710339 1'898299 12 1'425761 1'601032 4795856 2'012196 13 1'468534 1665074 1]885649 2'132928 14 1'512590 1'731676 i 1'979932 2'260904 15 1'557968 1-800944 2'078928 2'396558 16 1'604707 1'872981 2'182875 2'540352 17 1'652848 1'947900 2'292018 2'692773 18 1-702433 2025817 2'406619 2'854339 19 1'753506 2'1 06849 2'526950 3'021599 20 1'806111 2191123 2'653298 3'207135 21 1-860295 2278768 2'785963 3'399564 22 1'916103 i 2369919 2'925261 3'603537 23 1'973587 2'464716 3'071524 3'819750 24 2'032794 2'563304 3'225100 4'048935 25 2'093778 2'665836 3'386355 4'291871 26 2'156591 2'772470 3'555673 4'549383 27 2'221289 2'883369 3'733456 4'822346 28 2'287928 2'998703 3'920129 5-111637 29 2'356566 3'118651 4'4116136 5'418388 30 2'427262 3'243398 4'321942 5'743491 154 IUGIIER ARITH1M1IETiC. We will now solve the following questions by means of the preceding table. 4. What is the amount of $790 for 13 years, at 6 per cent.? From our table, we find the amount of $1 for 13 years, at 6 per cent., to be $2'132928; this, multiplied by 790, the number of dollars in the principal, gives $1685'013 for the amount required. 5. What is the compound interest of $49, for 20 years, at 5 per cent.? In this example, we find, from the table, that the amount of $1 for 20 years, at 5 per cent., is $2'653298; which, multiplied by 49, gives $130'012 for the amount of $49, from which, if we subtract $49, we get $81'012 for the compound interest required. 6. What is the compound interest of $100 for 17 years, at 6 per cent.? Ans. $169'277. 7. What is the conmpound interest of $375 for 20 years, at 6 per cent.? Ans. $827'676. 8. What is the amount of 8875 for 12 years, at 6 per cent., compound interest? Ans. $1760'672. 9. What is the amount of $625 for 18 years, at 5 per cent., compound interest? Ans. $81504'137. 10. What is the amount of $379 for 30 years, at 3 per cent., compound interest? Ans. $919'932. NOTE.-When the interest is compounded half-yearly, we must take the amount of $1 for half a year, and raise it to a power denoted by the number of half-years in the whole time; this power, multiplied by the principal, will give thle amount. We must proceed in a similar way for any other aliquot part of a year. Or, in such cases, we may make use of our table, as in the work of next question. 11. What is the amount of $100 for 3 years, at 6 per UOMPOUND DISCOUNT. 155 cent. per annum, when the interest is added at the end of every 6 months? In this example, we change the 6 per cent. to 3 per cent., and the 3 years to 6 years; we then find the tabular number to be $1'194052; which, multiplied by 100, gives $119'405 for the amount required. 12. What will ~600 amount to in 6 years, at 8 per cent., compound interest, supposing the interest to be receivable half-yearly? Ans. ~960 12 s. 4 — d. 13. What will $890 amount to in 5 years and 4 months, at 9 per cent. per annum, compound interest, the interest being added at the end of every 4 months? Ans. $1428189. 14. What will $3705 amount to in 3 years and 3 months, at 12 per cent. per annum, compound interest, the interest being added at the end of every 3 months? Ans. $5440'918. 15. What will $378 amount to in 7 years and 6 months, at 8 per cent. per annum, the interest being compounded half-yearly? Ans. $680'757. 16. What will $1000 amount to in 15 years, at 8 per cent. per annum, the interest being compounded halfyearly? Ans. $3243'398. COMPOUND DISCOUNT. 69. COMPOUND DISCOUNT is an allowance made for the payment of money before it is due, on the supposition that the money draws compound interest. The present worth of a debt payable at some future period, without interest, is such a sum as being put out 156 IHIIGIIER ARITHIMETIC. at compound interest, will, in the given tilne, at the given rate per cent., amount to the debt. Hence, the finding the present worth resolves itself into the following: Given the amount at compound interest, the time, and the rate per cent., to find the principal. Under compound interest, it was shown that the amount was equal to the number of dollars in tile principal, multiplied by the amount of $1 for one year, raised to a power whose exponent is the number of years. Hence, we have the following rule to find the principal, or present worth: R U L E. Divide the given amount by the amount of $1 for 1 year, raised to a power whose exponent is equal to the number of years. EX A M P I, S. 1. What is the present worth of $1685, due 13 years hence, allowing discount according to 6 per cent., compound interest? From the table under Art. 68, we find that the amount of $1 for 1 year, at 6 per cent., raised to tlhe 13th power, or, what is the same, the number for 13 years is $2'132928;.'. dividing $1685 by 2'132928, gives $789'994 for the present worth required. The present worth of $1 for one year, at 3 per cent., is r for two years, it is I is 0-; for two years, it is ); for three years, it is (.o 3)3; for four years, it is (.'O3)'; and in general, the present worth of $1, at compound interest, is the reciprocal of the amount of $1, at compound interest, for the same time and same rate per cent. UOMPOtUND) DISCOUNT. 157 TABLE, SHOWING the PRESENT WORTH of $1, or ~1, for any number of years, from 1 to 30, at 3, 4, 5, and 6 per cent., compound discount. Years.) 3 per cent. 4 per cent. 5 per cent, i 6 per cent. 1 0'970874 0'961538 0'952381 0'943396 2 0'942596 0'924556 0 907029 0'889996 3 0'915142 0'888996 0'863838 0'839619 4 0'888487 0854804 0'822702 0'792094 5 0'862609 0'821927 0'783526 0'747258 6 0'837484 0'790315 0'746215 0'704961 7 0'813092 0'759918 0710681 0'665057 8 0'789409 0'730690 0'676839 0'627412 9 0-766417 0-702587 0'644609 0'591898 10 0'744094 0'675564 0'613913 0'558395 11 0'722421 0'649581 0'584679 0'526788 12 0'701380 0'624597 0'556837! 0'496969 13 0-680958 0 600574 0'530321 0'468840 14 0'661118 0'577475 0'505068 0'442301 15 0'641862 0'555264 01481017 0'417265 16 0'623167 0'533908 0'458112 0'393646 17 0'605016 0'513373 0'436297 0'371364 18 0'587395 0'493628 0'415521 0'350344 19 0'570286 0'474642 0'395734 0'330513 20 0'553676 0'456387 0'376889 0'311805 21 0'537549 0'438834 0'358942 0'294155 22 0'521892 0'421955 0'341850 0'277505 23 0'506692 0'405726 0'325571 0'261797 24 0'491934 0'390121 0'310068 0'246979 25 0'477606 0'375117 0'295303 0'232999 26 0'463695 0'360689 0'281241 0219810 27 0'450189 0'346817 0'267848 0'207368 28 0'437077 0'333477 0'255094 0 195630 29 0'424346 0'320651 0'242946 0'184557 30 0'411987 0'308319 0'231377 0'174110 14 158 IIIG ER I ARITIhMETIC. The present worth of a given sum of money, discounting at compound interest, is easily obtained by the preceding table. 2. How much money must be placed out at compound interest to amount to $1000 in 20 years, the interest being 5 per cent.? Ans. $376'889. 3. What is the present worth of $1000, due 27 years hence, discounting at 3 per cent., compound interest? From the preceding table, we find the present worth of $1 for 27 years, at 3 per cent., to be $0'450189; this, multiplied by 1000, gives $450 189 for the present worth required. 4. What is the present worth of $3525, due in 3 years, discounting at 6 per cent., compound interest? Ans. $2959'657. 5. What is the present worth of $350, due 5 years hence, discounting at 6 per cent., compound interest? Ans. $261'54. 6. What is the present worth of $375, due 17 years hence, discounting at 4 per cent., compound interest? Ans. $192'515. 7. What is the present worth of $672, due 13 years hence, discounting at 5 per cent., compound interest? Ans. $356'376. 8. What is the present worth of $400, due 19 years hence, discounting at 6 per cent., compound interest? Ans. $132'205. 9. What is the present worth of $111, due 29 years hence, discounting at 3 per cent., compound interest? Ans. $47'102. ANNUITIES. 159 ANNUITIES. 70. AN ANNUITY is a fixed sum of money, which is paid periodically for a certain length of time. Case 1. To find the amount of an annuity which has been forborne for a given time. It is obvious that the last year's payment will be simply the annuity without any interest; the last but one will be the amount of the annuity for one year; the last but two will be the amount of the annuity for two years, and so on; and the sum of all these partial amounts will give the total amount due. Now we discover that these partial amounts, or payments, form a geometrical progression, whose first term is the annuity, the ratio is the amount of *1 for 1 year, and the number of terms is equal to the number of years; therefore, the amount of an annuity is found by summing the terms of a geometrical progression, when the first term, the number of terms, and the ratio, are-given. This may be done by the following RULE. From the amount of $1 for 1 year, raised to a power whose exponent is equal to the number of years, subtract $1; divide the remainder by the interest of $1 for 1 year; then multiply the annuity by this quotient. NOTE. —The different powers of the amount of $1 for one year, may be taken from the table under Art. 68. 160 IIGHER ARItTIIMETIC, EXAMPLES. 1. What is the amount of an annuity of $200, which has been forborne 14 years, at 6 per cent., compound interest? From table under Art. 6S, we find the 14th power of the amount of $1 for one year, at 6 per cent., to be $2' 260904; subtracting $1, and dividing the remainder by $0'06, the interest of $1 for one year, we get 21'01506; then multiplying $200, the annuity, by 21'01506, we find $4203'012 for the amount required. 2. Suppose a person, who has a salary of $700 a year, payable quarterly, to allow it to remain unpaid for 4 years; how much would be due him, allowing quarterly compound interest, at 12 per cent. per annum? Ans. $3527'453. 3. WVhat is due on a pension of $150 a year, payable half-yearly, but forborne 2 years, allowing half-yearly compound interest, at 6 per cent. per annum? Ans. $313'772. 4. What is due on a pension of $350 a year, payable quarterly, but forborne 24 years, allowing quarterly compound interest, at 12 per cent.? Ans. $1003'088. Questions under this rule may be easily wrought by the following table, which shows the amount of an annuity of $1, or ~1, forborne for any number of years not exceeding 30, at 3, 4, 5, and 6 per cent., compound interest. ANNUITIES. 1G1 TABLE OF ANNUITIES. Years. 3 per cent, per cent. 5 per cent. 6 per cent. 1'000000 1'000000 1 1000000 1'000000 2 2'030000 2'040000 2'050000 2'060000 3 3'090900 3'121600 3'152500 381S3600 4 4183627 4'246464 4'310125 4'374616 5 5'309136 5'416323 5'525631 5'637093 6 6'468410 6 632975 6'801913 6'975319 7 7'662462 7'898294 8'142008 8'393838 8 8'892336 9'514226 9'249109 9'897468 9 10'159106 1 0'582795 1 1'026564 11'491316 10 11'463879 12'006107 12.577893 13'180795 11 12'807796 1 1348 6331 14'206787 14'971643 12 14192030 015'025805 15'917127 16'869941 13 15'617790 16'626838 17'712983 18'882138 14 1708324 18'291911 19'598632 21'015066.15 18'5980V14 20'023588 21'578564 23'275970 16 20-156881 21-824531 23'657492 25'6725f28 17 21'761588 293'697512 25'840366 28'212880 18 23'414435 25'645431 28'132385 30'905653 19 25'116868 27'671229 30'539004 33'759992 20 26'870374 29'778079 33'065954 36'785591 21 28'676486 31'969202 35'719252 39'992727 22 30'536780 34'247970 38'505214 43'392290 23 32'452884 36'617889 41'430475 46'995828 24 34'426470 39'082604 44'501999 50'815577 25 36'459264 41.645908 47727099 54'864512 26 38'553042 44.311745 51'113454 59'156383 27 40'709634 47'084214 54'669126 63'705766 28 42'930923 49'967583 58'402583 68'528112 29 45'218850 52'966286 62'322712 73'639798 30 47'575416 560'084938 66'438847 79'058186 162 HIIGHIER ARITHME'TItC. 5. What is due on a pension of $1000, which has been forborne 27 years, at 3 per cent., compound interest? From the preceding table, we find the amnount of an annuity of $1 for 27 years, at 3 per cent., to be $40 709634, which, multiplied by 1000, gives $40709 634 for the amount due. 6. What is the amount of an annuity of $50, which has been forborne 30 years, at 6 per cent., compound interest? Ans. $3952 909. 7. What is the amount of a pension of $300, which has been forborne 19 years, at 5 per cent., compound interest? Ans. $9161'701. 8. What is the amount of a pension of $900, which has been forborne 17 years, at 4 per celt., compound interest? Ans. $21327'761. 9. What is the amount of an annuity of $75, which has been forborne 13 years, at 5 per cent., compound interest? Ans. $1328'474. Case 11. To find the present worth of an annuity which is to terminate in a given number of years. The present worth of an annuity is obviously such ia sum of money as will, at compound interest, produce an amount equal to tile amount of the annuity. Therefore, if we find the amount of the annuity by Case I., we may consider it as the amount of a certain principal, which principal is the same as the present worth. We have already been taught how to find the present worth, by rule under Compound Discount. Hence, we have this ANNUITIES. 163 RULE. First find the amount of the annuity, as if it were in arrears for the whole time, by the aid of the table under Case I. of ANNUITIES. Then find the present worth of this amount for the given time and rate per cent., by the use of the table under COMPOUND DISCOUNT. EXAMPLES. 1. What is the present worth of an annuity of $500, to continue 10 years, interest being 6 per cent.? By the table under Case I., of Annuities, we find the amount of an annuity of $1 for 10 years, at 6 per cent., to be $13'180795; this, multiplied by 500, gives $6590'3975 for the amount of the annuity. Now, by the table under Compound Discount, we find the present worth of $1 for 10 years, at.6 per cent., to be $0'558395; which, multiplied by 6590'3975, gives $3680'045 for the present worth required. 2. What is the present worth of an annuity of $100, to continue 20 years, at 5 per cent. interest? Ans. $1246'222. The work under this rule may be very much simplified by the use of the following table, which gives the present worth of an annuity of.$1, or ~1, for any number of years not exceeding 30, at 3, 4, 5, and 6 per cent. 164 BHIGHER ARITHMETIC. TABLE OF ANNUITIES. Years. 3 per cent. 4 per cent. 5 per cent. 6 per cent. 1 0'970874 0'961538 0'952381 0'943396 2 1'913470 1'086095 1'859410 1'833393 3 2'828611 2'775091 2'723248 2'673012 4 3'717098 3'629895 3'545950 3'465106 5 4'579707 4'451822 4'329477 4'212364 6 5'417191 5'242137 5'075692 4'917324 7 6'230283 6'002055 5'786373 5'582381 8 7'019692 6'732745 6'463213 6'209794 9 7'786109 7'435332 7'107822 6'801692 10 8'530203 8'110896 7'721735 7'360087 11 9'252624 8'760477 8'306414 7'886875 12 9'954004 9'385074 8:863252 8'383844 13 10'634955 9'985648 9'393573 8'852683 14 11'296070 10'563122 9'898641 9'294984 15 11'936935 11'118387 10'379658 9'712249 16 12'561102 11'652296 10'837770 10'105895 17 13'166118 12'165669 11'274066 10'477260 18 13'753513 12'659197 11'689587 10'827603 19 14'323799 13'133839 12'085321 11'158116 20 14'877475 13'590326 12'462216 11'469921 21 15'415024 14'029160 12'821153 11'764077 22 15'936917 14'451115 13'163003 12'041582 23 16'443608 14'856842 13'488574 12'303379 24 16'935542 15'246963 13'798642 12'550358 25 17'413418 15'622080 14'093945 12'783356 26 17'876842 15'982769 14'375185 13'003166 27 18'327031 16'329586 14'643034 13'210534 28 18'764108 16'663063 14'898127 13'406164 29 19'188455 16'983715 15'141074 13'590721 30 19'600441 17'292033 15'372451 13'764831 ANN UITIES. 165 To find the present worth of an annuity by means of this table, we must take from it the present worth of $1 for the given time and rate per cent., and multiply it by the number of dollars in the given annuity. 3. What is the present worth of an annuity of $27 for 9 years, at 4 per cent.? From the table, we find the present worth of $1 for 9 years, at 4 per cent., to be $7'435332; this, multiplied by 27, gives $200'754 for the pres6nt worth. 4. What is the present worth of a pension of 875 for 15 years, at 5 per cent.? Ans. $778'474. 5. A young man purchases a farm for $924, and agrees to pay for it in the course of 7 years, paying - part of the price at the end of each year. Allowing interest to be 6 per cent., how much cash in advance will pay the debt? Ans. $736'874. 6. Allowing interest to be 6 per cent., how much shall I gain by paying $15 a year for 10 years, in order to cancel a debt of $160, now due? Ans. $49'599. 7. What is the present worth of an annuity of $375 for 13 years, interest being reckoned at 4 per cent.? Since 375 is 3 of 1000, we may multiply the tabular number by 3, divide by 8, and remove the decimal point three places to the right. Oporation. 9'985648 3 8)29'956944 Ans. $3744'618 8. What is the present worth of an annuity of $875 for 11 years, interest being 6 per cent.? 166 HIGHER ARITHMETrTC. Since 875 is 7 of 1000, we may multiply by 7, divide by 8, and remove the decimal point three places to the right. Operation. 7'886875 7 8)55'208125 Ans. $6901'015625 Or, we might have subtracted - of the tabular number from itself, and then have removed the decimal point three places to the right, as in this second Operation. 8)7'886875 985859375 Ans. $6901l015625 The student ought to exercise himself in seeking short and expeditious methods whenever the nature of the operation will admit of contractions. NOTE.-When an annuity does not commence until a given time has elapsed, or some particular event has happened, it is called a REVERSION. Case III. To find the present worth of an annuity in reversion. RULE. Find, by the use of the table under last Case, the present worth of the annuity from the present time up to the end of its continuance; fnd, also, by the same table, its value for the time before it commences; the diffcrcnce cf these results will be the present worth. ANNUITIES. 167 Or, which is the same thing: Take the difference of the tabular numbers for these two periods, and multiply by the number of dollars in the annuity. EXAMPLES. 1. What is the present worth of an annuity of $200, to be continued 5 years, but not to commence till 2 years hence, interest being 6 per cent.? By ollr table, we find the present worth of $1 for 7 years, at 6 per cent., to be $5'582381; the same for 2 years, is $1'833393; the difference is $3'748988, which, multiplied by 200, gives $749'798 for the present worth. 2. A father leaves to his son a rent of $310 per annum for S years, and the reversion of the same rent to his daughter for 14 years thereafter. What is the present worth of the legacy of each, at 6 per cent.? Operation. 6'209794 —tabular number for 8 years. 310 62097940 18629382 $1925'036140 —present worth of son's. 12'041582 - tabular number 8 + 14 = 22 years. 6'209794=tabular number for 8 years. 5'831788 310 58317880 17495364 $1807'854280 = present worth of daughter's. 3. What is the present worth of a reversion of $100 168 H IGHEIR ARITIINMETIC. a year, to commence in 4 years, and to continue for 10 years, interest being at 6 per cent.? Ans. $58298S8. 4. What is the present worth of a reversion of $800 a year, to continue 7 years, but not to commence until the end of 8 years, interest being 4 per cent.? Ans. $3508'514. When the annuity is to continue for ever, it is obvious that its present worth will be that sum whose interest for 1 year is equal to the annuity; therefore, to find the present worth of an annuity to continue for ever, we must divide the annuity by the interest of $1 for 1 year, at the given rate per cent. 5. How much must be paid at present for the title to an annuity of $1000, to commence in 7 years, and to continue for ever, interest at 6 per cent.? Dividing $1000 by $0'06, we get, for the present worth, if entered tipon imnmediatcly,,$1666666. From table under Compound Discoullt, we findl the present worth of $1 for 7 years, at 6 per cent., to be $0'665057; this, multiplied by 16666'66, gives 811084'283 for the present worth of $166666'663, which is evidently the same as the present worth of the annuity. 6. What is tlhe present worth of a reversion of $100 a year, to commieonce in 4 years, and to continue for ever, interest being 6 per cent.? Dividing $100 by $0'06, we get 16662- for the number of dollars in the present worth, if entered upon immediately. From the table under Compound Discount, we find the present worth of $1 for 4 years, at 6 per cent., to be $0'792094, which must be multiplied by 16662. But, since 16662 is - of 1000, we nmay multiply by 5, divide ANNUIT lES. 169 by 3, and remove the decimal point three places to the right, as in the following Operation. $0'792074 5 3)3960470 Ans. $1320'153 71. The following tables, which have been computed by the aid of logarithms, are added more for curiosity than for any view to their utility. This table gives the time required for a given principal to double itself, at compound interest, the interest being compounded yearly. Per cent. Years. Per cent. Years. Per cent. Years. 1 69'666 4 17 673 7 10'245 1 ~ 46'556 4: 15I748 72 9'585 2 35'004 5 14 207 8 9'006 2{ 28'071 51 12 946 8-1 8'497 3 23'450 6 1'896 9 8'043 32 20'150 6! 1'007 9f 7'638 The following table gives the time required for a* given principal to double itself at compound interest, the interest being compounded half-yearly. 15 170 HIGHER ARITHMETIC. Per cent. Years. Per cent. Years. Per cent. Years. 1 69'487 4 17'502 7 10'075 1- 46'382 41 15'576 71 9'914 2 34-830 5 14'036 8 8'837 21 27-899 5- 12'775 8- 8'346 3 23'278 6 1 1725 9 7'874 3t 119'977 6- 10'836 9- 7'468 This table gives the time required for a given principal to double itself at compound interest, the interest being compounded quarter-yearly. Per cent. Years. Per cent. Years. Per cent. Years.1 69'400 4 17'415 7 9'989 1- 46'298 40- 15'490 71 9'328 2 34'743 5 13'946 8 8'751 21 27'812 51 12'686 8- 8'241 3 23'191 6 11l639 9 7'788 31- 19'890 6'- 10'750 91 7'383 The following table gives the time required for a given principal to double itself at compound interest, the interest being compounded every instant. Per cent. Years. Per cent. Years. Per cent. 1 69'315 4 17'329 7 9'902 1{ 46'210 4- 15'403 7- 9'242 2 34'657 5 13'863 8 8'665, 2 27'726 5 12'603 8f 8'15.5 3 23'105 6 1'1552 9 7'702 31 19'804 6- 10'664 9) 7'296 INSTANTANEOUS INTEREST. 171 The following table gives the amount of $1, or ~1, for any number of years up to 30, for 5 and 6 per cent., compound interest, the interest being compounded every instant. Years. 5 per cent. 6 per cent. Years. 5 per cent. 6 per cent. 1 1 0513 1 0618 16 2'2255 2'6116 2 1 1052 1 1275 17 2'3396 2-7731 3 1161S 1-1972 18 2'4595 2'9446 4 1-2214 1-2712 19 2-5857 3'1267 5 1-2840 1 3498 20 2'71'82'33201 6 13498 1P4333 21 2'8576 3'5253 7 1 4190 1 5219 22 3'0041 3'7433 8 1-4918 1 5161 23 3 1581 3'9748 9 1'5683 1-7160 24 333201 4'2206 10 1'6487 18221 25 3 4903 4'4815 11 1P7332 1 9.348 26 3'6693 4'7587 12 1 8221 2-0544 27 3 8573 5'0529 13 1'9155 2 1771 28 4 0550 5'3653 14 220137 2'3163 29 4 2630 5'6971 15 21 169 244596 30 4:4815 6'0492 If we compute the instantaneous compound interest at 6'76587 per cent., it will, at the end of the year, be equal to the simple interest at 7 per cent. In the same way, the instantaneous compound interest, at 5'8269 per cent., is the same as simple interest at 6 per cent. [For some curious results in regard to Instantaneous Compound Interest, see an article which I prepared for the American Journal of Science, Vol. 47, No. 1.] 172 HIGHER ARITHMETIC. CHAPTER VIII. BANKING. 72. A BANK is an incorporated institution, created for the purpose of loaning money, receiving deposits, and dealing in exchange. The Stock, or amount of money in trade, is limited by law, and owned by various individuals, who are called stockholders. Banks are allowed to make notes, which are denominated bank bills, which circulate as money, because they are obliged to redeem them with specie. It is customary for banks, in most cases, when they loan money, to take the interest in advance;* that is, to deduct it from the face of the note at the time the money is lent. The note is then said to be discounted. The sum to be discounted, or the face of the note, is called the amount. The interest deducted is called the discount. What remains is called the present worth, or proceeds. A note to be discounted, or bankable, must be made payable at some future time, and to the order of some person who indorses it. It is usual for the banks to take interest for three days more than the time specified in the note; and the borrower is not obliged to make payment till those three * This method of discounting bank notes is usurious, and is fast going out of use, and instead thereof the banks now deduct the discount as found by rule under Art. 66. BANKING. 173 days have expired, which are, for this reason, called days of grace. To find the banking discount on any sum of money, we have this RULE. Compute the interest (by Case III., Art. 63,) on the given sum for three days more than is specified. EXAMPLES. 1. What is the banking discount on $1000 for 3 months, at 7 per cent.? In this example, we find the interest on $1 for 3 months and 3 days, at 6 per cent., to be $0'0155, which, multiplied by 1000, gives $15'50 for the discount at 6 per cent.; this, increased by its sixth part, becomes $18'08} for the discount at 7 per cent., as required. 2. What is the banking discount of $150 for 6 months, at 6 per cent.? Ans. $4'575. 3. What is the banking discount of $375 for 3 months and 9 days, at 7 per cent.? Ans. $7'438. 4. What is the banking discount of $400 for 9 months, at 7 per cent.? Ans. $21'233. 5. What is the banking discount of $29'30 for 7 months, at 5 per cent.? Ans. $0'867. 6. What is the banking discount of $472 for 10 months, at 7 per cent.? Ans. $27'808. When the present worth of a bankable note, the time for which it is to be discontinued, and the rate per cent. is given, to find the amount, we have this RULE. Compute the banking discount on $1 for the given time and rate per cent.; subtract this discount from $1, 15* 174 IIIGIIER ARITHMETIC. then divide the present worth by the remainder, and the quotient will be the number of dollars in the amount. EXAMPLES. 1. What must be the amount of a bankable note, so that, when discounted for 3 months, at 6 per cent., it shall give a present worth of $600? In this example, we find the banking discount on $1 for 3 months, to be $0'0155, which, subtracted from $1, gives $0'9845;.'. dividing $600 by $0'9845, we obtain 609'446 for the number of dollars in the required amount of the note. 2. What must be the face of a bankable note, so that, when discounted for 2 months, at 7 per cent., the borrower shall receive $50? Ans. $50'62. The following table gives the amount of a bankable note, so that, when discounted at 5, 6, or 7 per cent., for any number of months from 1 to 12, the present worth shall be just $1. Months. 5 per cent. 6 per cent. 7 per cent. 1 1'004604 1'005530 1'006458 2 1'008827 1 010611 1'012402 3 1'013085 1'015744 1'018416 4 1P017380 1'020929 1'024503 5 1P021711 1026167 1'030662 6 1P026079 1P031460 1P036896 7 1 030485 1'036807 1P043206 8 1P034929 1P042095 1'049593 9 1 039411 1 047669 1P056059 10 1 043932 1'053186 1P062605 11 1P048493 1'058761 1'069233 12 1P053093 1P064396 1P075944 BANKING. 175 We will now work some examples by the aid of the preceding table: 3. What must be the face of a bankable note, so that, when discounted for 10 months, at 5 per cent., the present worth may be $1000? Looking in the table, directly under the 5 per cent., and adjacent to 10 months, we find $1-043932; this, multiplied by 1000, gives $1043-932 for the face of the note required. 4. What must be the face of a bankable note, so that, when discounted for 7 months, at 7 per cent., the present worth may be $70 50? Ans. $73x546. 5. What aniount must I make my note, so that, when discounted at the bank for 12 months, at 7 per cent., I may receive $100? Ans. $107-594. 6. What must be the amount of a note, so that, when discounted at the bank for 6 months, at 6 per cent., the borrower may receive $365? Ans. $376-483. 7. What must be the amount of a note, so that, when discounted at the bank for 9 months, at 7 per cent., the borrower may receive $500? Ans. $527'03. 73. The banks, by this method of discounting, obtain a larger per cent. for their money than is obtained by the usual method of loaning money. To illustrate this, suppose A gets a note of $1 discounted at the bank for 12 months, or 1 year, at 7 per cent., he receives $0'93; the $0'07 is retained by the bank, it being the interest 176 H 1 f IIE R A R ITHT EM i C. of $1 for 1 year. This $0'07 may now be loaned to B, and its interest again withheld; and so on, for an indefinite period of terms. Hence, at the end of the year, the bank will receive for its $1, the number of dollars expressed by the sum of the terms of the following geometrical progression: 1 +i _+(T- )2 +(-X7 )3 +, &c.; this, summed, disregarding the 3 days of grace, gives -o_- =1'0752688. Therefore, in this case, the bank receives 7'52688 per cent. per annum for its money. The longer the time for which they discount, the larger per cent. do they receive. To make this appear obvious, suppose a person wished his note discounted at the bank for 144- years, at 7 per cent. In this case, the interest would equal the whole face of the note; so that the bank would withhold the whole amount, be that ever so large, and the borrower would not receive a single cent, but would, nevertheless, be obliged to pay to the bank, at the end of 144 years, the face of the note. In this case, the per cent. would be infinite. If we go one step farther, and endeavor to discount a note at the bank for a longer period than 142 years at 7 per cent., we shall be obliged to pay to the bank money from our own pocket before they would accept our note. The following table shows the per cent. received by banks, when their notes are renewed at the end of any number of months from 1 to 12, at 5, 6, and 7 per cent., lawful interest. RANKING 177 Months. 5 per cent. 6 per cent. 7 per cent. 1 5'138 6'200 7'272 2 5'149 6'216 7'295 3 5'160 6'232 7'317 4 5'1 72 6'248 7'339 5 5'183 6'264 7'362 6 5' 194 6'281 7'385 7 5'205 6'298 7'408 8 5'217 6'315 7'432 9 5'228 6'332 7'456 10 5'239 6'349 7'480 11 5'251 6'366 7'503 12 5'263 6'383 7'527 NOTE.-Were it possible for banks to renew their notes every instant, the respective rates per cent. would be 5'127, 6-182, and 7'251. This is the same as would be received if the interest were added every instant. 178 1HIGIIER AIR 1111METIC. CHAPTER IX. INVOLUTION. 74. INVOLUTION is the method of finding the powers of numbers. We have already defined the power of a number to be the result arising from multiplying it into itself continually, until the number has been used as a factor as many times as there are units in the exponent denoting the power. Thus, to obtain the cube, or third power of 7, we must use it as a factor three times, which will produce 7 x 7 x 7= 343. EXAMPLES. 1. What is the square of 23? Ans. 529. 2. What is the cube of 17? Ans. 4913. 3. What is the 5th power of 47? Ans. 229345007. 4. What is the 9th power of 9? Ans. 387420489. 5. What is the square of 22667121? Ans. 513798374428641. 6. What is the square of 0'75'? Ans. 0'5625. 7. What is the cube of 0'65? Ans. 0'274625. 8. What is the square of 8}? Ans. 72~. EXTRACTION OF THE SQUARE ROOT. 179 EVOLUTION. 75. EVOLUTION is the reverse of Involution. It explains the method of resolving a number into equal factors, which factors are called roots. When a number is resolved into two equal factors, this factor is called the square root of the number. When a number is resolved into three equal factors, the factor is called the cube root of the number. The operation of resolving a number into two equal factors is called the extraction of the square root. EXTRACTION OF THE SQUARE ROOT. 76. IF we square 48 by the usual rule, we get 482=2304. But if, instead of 48, we use 40+8, we shall find, by actual multiplication, 40+8 40+8 320+ 64 1600+320 482 = 1600 + 640 + 64 Now, to reverse this operation, that is, to extract the square root of 1600 + 640 + 64, we proceed as follows: We take the square root of 1600, which is 40; this is the first part of the root; its square being subtracted from 1600+640-+64, leaves the remainder 640+64. We see that 640, divided by twice 40, or 80, gives 8 for a quotitent, which is the second part of the root required. 180 HIGHER ARITHMETIC. Case L From the preceding process, we deduce the following rule for the extraction of the square root of a whole number: RULE. I. Point of the given number into periods of two figures each, counting from the right towards the, left. When the number of figures is odd, it is evident that the left-hand, or first period, twill consist of but one figure. II. Find the g'reatest square in the first period, and place its root at the right of the number, in the form of a quotient figure in division. Subtract the square of this root from the first period, and to the remainder annex the second period; the result will be the FIRST DIVIDEND. III. Double the root already Jound, and place it on the left of the number for the FIRST TRIAL DIVISOR. See how many times this trial divisor, with a cipher annexed, is contained in the dividend; the quotient figure will be the second figure of the root: this must be placed at the right of the TRIAL DIVISOR; the result will be the TRUE DIVISOR. Multiply the true divisor by this second figure of the root, and subtract the product from the dividend, and to the remainder annex the next period for a SECOND DIVIDEND. IV. To the last TRUE DIVISOR add the last figure of the root for a new TRIAL DIVISOR, and continue to EXTRACTION OF THE SQUARE ROOT. 181 operate as before, until all the periods have been brought down. EXAMPLES. 1. What is the square root of 531441? Operation. 53'14'41(729, root. 7 49 First trial divisor, 14 414, first dividend. First true divisor, 142 284 Second trial divisor, 144 13041, second dividend. Second true divisor, 1449 13041 0 2. What is the square root of 11390625? Operation. 3 11'39'06'25)3375 63 9 667 239 6745 189 5006 4669 33725 33725 0 In the first example, we exhibited the trial divisors, as well as the true divisors; but in the second example, we adhered more closely to our rule, and placed the succeeding figures of the root at the right of the trial divisors, without again writing them down. 16 182 HIGHER ARITHIMETIC. 3. What is the square root of 11019960576? Operation. 1 1'10'19'96'05'76(104976 204 1 3089 1019 20987 816 209946 20396 18801 159505 146909 1259676 1259676 0 4. What is the square root of 16983563041? Ans. 130321. 5. What is the square root of 79792266297612,001? Ans. 282475249. 6. What is the square root of 852891037441? Ans. 923521. 7. What is the square root of 61917364224? Ans. 248832. 8. What is the square root of 13422659310152401? Ans. 115856201. 9. The sum of the four numbers, 386, 2114, 3970, 10430, is a perfect square; so also is the sum of any two of these numbers. What are these seven roots? Ans. 130, 50, 66, 104, 78, 112, 120. Case II. To extract the square root of a decimal fraction, or of a number consisting partly of a whole number, and partly of a decimal value, we have this EXTRACTION OF THE SQUARE ROOT. 183 RULE. I. Annex one cipher, if necessary, so that the number of decimals shall be even. II. Then point off the decimals into periods of two figures each, counting fJom the unit's place towards the right. If there are whole numbers, they must be pointed off as in Case I. Then extract the root as in Case I. NoT-E.-If the given number has not an exact root, there will be a remainder after all the periods have been brought down, in which case the operation may be extended by forming new periods of ciphers. EXAMPLES. 1. What is the square root of 3486'784401? Ans. 59'049. 2. What is the square root of 25'62890625? Ans. 5'0625. 3. What is the square root of 6'5536? Ans. 2'56. 4. What is the square root of 0'00390625? Ans 0'0625. 5. What is the square root of 17? Ans. 4'123, nearly. 6. What is the square root of 37'5? Ans. 6'123, nearly. 7. What is the square root of 0'0000012321? Ans. 0'00111. 8. What is the square root of 0'0011943936? Ans. 0'03456. 9. What is the square root of 60'481729? Ans. 7'777. Case III. To extract the square root of a vulgar fraction, or mixed number, we have this 184 HIGHER ARITHMETIC. RULE. I. Reduce the vulgar fraction, or mixed number, to its simplest fractional form. II. Then extract the square root of the numerator and denominator separately, when they have exact roots; but when they have not, reduce the fraction to a decimal, and proceed as in Case II. EXAMPLES. 1. What is the square root of 2 5? Ans.. 2. What is the square root of o01-5_? Ans. 45 3. What is the square root of 42-? Ans. 2~. 4. What is the square root of 5 of 4 of 4 of? Ans. 4. 5. What is the square root of 49? Ans. 2'027, nearly. 6. What is the square root of -7? Ans. 0'8044, nearly. 7. What is the square root of -9? Ans. 0515, nearly. S. What is the square root of ~3? Ans. 0'052, nearly. 9. What is the square root of.234? Ans. 0'534, nearly. 10. What is the square root of 24 4 4? Ans. 0-524, nearly. Case IV. When there are many figures required in the root, we may, after obtaining one more than half the number required, find the rest by dividing the remainder by the last true divisor, deprived of its right-hand figure. This division should be performed according to the abridged method, as explained under Art. 41. EXAMPLES. 1. What is the square root of I to 16 decimals? EXTRACUTION OF THE SQUARE ROOT. 185 Operation. 3 11(3'3166247903553998. 63 9 661 2 6626 189 66326 11 663322 661 6633244 439 66332487 39756 66332494[9 4144 397956 16444 1326644 317756 26532976 5242624 464327409 59934991 5969924541 23574559 19899748 3674811 3316625 358186 331662 26524 19900 6624 5970 654 597 57 53 1A, 186 HIIGHER ARITHIMETIC. In the preceding examnple, after obtaining 9 figures of the root, by the usual rule, we had, for the remainder, 23574559; the last true divisor was 6633249419, when deprived of its right-hand figure. We then divided this remainder by this divisor, according to the method of abridged divisions cf decimals, Art. 41, and obtained the remaining 8 figures of the root. 2. What is the square root of 3 to 10 decimals? Arns. 1'7320508076. 3. What is the square root of 0'00008876684 to 10 places of decimals? AAns. 0'00942161 55. 4. What is the square root of 0'8867081113724 to 10 places of decimals? Ans. 0'9416517994. 5. What is the square root of 3'14159265 to 8 places of decimals? Anzs. 1'77245385. 6. What is the square root of 2 to 9 places of decimals? Anrs. 1'414213562. 7. What is the square root of 10 to 15 places of decimals? Ans. 3 162277660168379. 8. What is the square root of the decimal 0'4444, &c., or of the simple repetend 0'4? Ans. 0'666, &c., or. From this example, we see that tile square root of a repetend may also be a repetend. EXAMPLES INVOLTVING THIE PRINCIPLES OF THE SQUARE ROOT. 77. A TRIANGLE is a figure having three sides, and, consequently, three angles. APPLICATION OF rtIE SQUARE ROOT. 187 When one of the angles is right, like the corner of a square, the triangle is called a r-ight-angled triangle. In this case, the side opposite the right angle is called the hypothenuse. It is an established proposition of geometry, that the square of the hypothenuse is equal to the sum of the squares of the other two sides. From the above proposition, it follows that the square of the hypothenuse, diminished by the square of one of the sides, equals the square of the other side. By means of these properties, when two sides of a right-angled triangle are given, the third side can be found. EXAMPLES. 1. How long must a ladder be to reach the top of a house 40 feet high, when the foot of it is 30 feet from the house? In this example, it is obvious that the ladder- forms the hypothenuse of a right-angled triangle, whose sides are 30 and 40 feet, respectively. Therefore, the square of the length of the ladder must equal the sum of the squares of 30 and 40. 302 - 900 402 - 1600 2/2500 =50=the length of the ladder. 2. Suppose a ladder 100 feet long be placed 60 feet from the foot of a tree; how far up the tree will the top of the ladder reach? Ans. 80 feet. 3. Two persons start from the same place, and go, the one due north 50 miles, the other due west 80 miles. How far apart are they? Ans. 94'34 miles, nearly. 188 HIGHER ARITHMETIC. 4. What is the distance through the opposite corners of a square yard? Ans. 4'24264 feet, nearly. 5. The distance between the lower ends of two equal rafters, in the different sides of a roof, is 32 feet, and the height of the ridge, above the foot of the rafters, is 12 feet. What is the length of a rafter? Ans. 20 feet, nearly. 6. What is the distance measured through the centre of a cube, from one corner to its opposite corner, the cube being 3 feet, or 1 yard, on a side? Ans. 5'196 feet. We know, from the principles of geometry, that all similar surfaces, or areas, are to each other as the squares of their like dimensions. 7. Suppose we have two circular pieces of land, the one 100 feet in diameter, the other 20 feet in diameter; how much more land is there in the larger than in the smaller? By the above principle of geometry, it follows that the quantity of land in the two circles must be as the squares of the diameters; that is, as 1002 to 202, or as 25 to 1. Hence, there is 25 times as much in the one piece as there is in the other. 8. Suppose two persons, the one 6 feet high, the other 5 feet, to be both well proportioned in all respects; how much more cloth will it take to make a suit of clothes for the first, than for the second? It will require 1 times as much for the Afirst as for the second. 9. Suppose, by observation, it is found that 4 gallons of water flow through a circular orifice of 1 inch in diameter, in one minute; how many gallons would, under similar circumstances, be discharged through an orifice of 3 inches in diameter, in the same length of time? Ans. 36 gallons. APPLICATION OF THE SQUARE ROOT. 189 10. What must be the circumference of a circular pond, which shall contain ~ — part as much surface as a pond 132 miles in circumference? Ans. 33 miles. 11. Required the width and depth of a rectangular box, whose length is 3 feet, which shall contain 30000 solid inches; the width being to the depth as 2 to 3. A box, whose length is 3 feet = 36 inches, width 2 inches, and depth 3 inches, must contain 36 x 2 x 3 inches. Then, 0 X,L X - X 3 l 29 0, whose square root is 2, V/2=11'785, nearly; this, multiplied respectively by 2 and 3, will give s 1 1 785 x 2= 23'57 inches. Ans. 11785 x 3-35'36 " Or, this question may be solved by the following nethod: Had the width of the box been the same as the depth, its volume would have been one half more than it now is; that is, would have been 2 of 30000-45000 cubic inches, which, divided by 36 inches, the length, will give 1250 square inches for the end area of this new box; or, which is the same thing, 1250 is the square of the depth of the original box. Hence, 1250 = 35'36, nearly, for the number of inches in the depth. The width is 2 of 35'36-23'57, nearly. 12. What length of thread is required to wind spirally around a cylinder 2 feet in circumference, and 3 feet in length, so as to go but once around? It is evident that if the cylinder be developed, or placed upon a plane, and caused to roll once over, that the convex surface of. the cylinder will give a rectangle, whose width is 2 feet, and length 3 feet; at the same 190 HIGHER ARITHMETIC. time the thread will form its diagonal. Hence, the length of the thread is 4 + 9- 13 —3' = 360555 feet. 13. Seven men purchase a grindstone, of 60 inches in diameter. What part of the diameter must each grind off, so as to have I of the whole stone? Solution. In this question, we disregard the thickness of the stone. After the first one has ground off his share, the remaining stone will be 6 of the original stone. Therefore, its diameter will be 60 V/J=-?- 4/~2=55'54921, nearly. The diameter, after the second one has ground off his share, will be 60 =/- 67/3 = 50'70925, nearly. The diameter, after the third one has ground off his share, will be 60V/; —J6- V2/8=45'35574, nearly. The diameter, after the fourth one has ground off his share, will be 60 V/=-6 — -/ = 39'27922, nearly. The diameter, after the fifth one has ground off his share, will be 60 A/ =-67Q- /14=32'07135, nearly. The diameter, after the sixth one has ground off his share, will be 60 /' --- 6-77= 22 67787, nearly. Hence, the parts of the diameter ground off are as follows: Inches, nearly. The 1st ground off 60'00000 —55'54921 = 4'45079 2d " " 55'54921-50'70925- 4'83996 3d " " 50'70925-45'35574= 5'35351 4th " " 45'35574-39'27922- 6'07652 5th " " 39'27922-32'07135- 7'20787 6th " " 32'07135-22'67787- 9'39348 7th " " 22'67787 =22 67787 EXTRACTION OF THIE CUBE ROOT. 191 EXTRACTION OF THE CUBE ROOT. 7S. IF we cube 45 by the usual process, we find 453 — 91125. If, instead of 45, we take its equal, 40+5, and then cube it by actual multiplication, as explained under Art. 4, we shall have this Operation. 45 —40+ 5 40+ 5 200+ 25 1600+ 200 452 =1600+ 400+ 25 40+ 5 8000 + 2000 + 125 64000+ 16000+ 1000 453 = 64000 + 24000 + 3000 + 125 Now, to reverse this process, that is, to extract the cube root of 64000 + 24000 + 3000 + 125, we proceed as follows: I. We find the cube root of 64000 to be 40, which we place to the right of the number, in the form of a quotient in division, for the first part of the root sought. We also place it on the left of the number in a column headed 1st COL.; we next multiply it into itself, and place the result in a column headed 2d COL.; this last result, being multiplied by 40, gives 64000, which we subtract from the number, and obtain the remainder 24000+3000+ 125, which we will call the first dividend. 192 HIGHER ARITHMETIC. II. We obtain the second term of the 1st column by adding the first term to itself; the result being multiplied by this first term, and added to the first term of the 2d column, gives its second term. Again, adding this first term to the second term of the 1st column, we get its third term. III. We seek how many times the second term of the 2d column is contained in the first dividend; or, simply how many times it is contained in its first part, 24000, whichlgives 5 for the second part of the root. IV. Finally, we add this 5 to the last termn of the 1st column, whose result, multiplied by 5, and added to the last term of the 2d column, gives its third term; which, multiplied by 5, gives 27125 - 24000 -t-3000+ 125. 1st COL. 2d COL. Number. Root. 40 1600 64000+24000+3000+ 125(40+5. 80 4800 64000 120 5425 24000+3000+ 125= 27125 125 5425 x 5 = 27125 This work can be written in a more condensed form, as follows, where the ciphers upon the right have been omitted. Number. Root. 1st CoL. 2d COL. 91125 ( 45 4 16 64 8 48 27125 12 5425 27125 125 0 EXTRACTION OF THE CUBE ROOT. 193 Case I. From the preceding operation, we may draw the following rule for extracting the cube root of a whole number. RULE. I. Since the cube of any number cannot have more than three times as many places of figures as the number, we must separate the number into periods of three figures each, counting from the unit's place towards the left. T4When the number of figures is not divisible by 3, the left-hand period will contain less than three figures. II. Seek the greatest cube of the first, or left-hand period; place its root at the right of the number, after the manner of a quotient in division; also place it to the left of the number for the first term of a column marked 1st CoL.'Then multiply it into itself, and place the product for the first term /f ra column marked 2d COL. Again, multiply this last result by the same figure, and subtract the product fromn the first period, and to the remainder annex the next perio(l, and it will give the FIRST DIVIDEND. This same figure must be added to the first term of the 1st column; the sum will be its second term, which must be multiplied by the same figure, and the product added to the first term of the 2d column; this sum will be its second term, which we shall name the FIRST TRIAL DIVISOR. The same figure of the root must be added to the second term of the 1st column, to form its third term, III. See how many times the trial divisor, with two 17 194 HIGHER ARITHMETIC. ciphers annexed, is contained in the dividend; the quotient figure will be the second figure of the root, which must be placed at the right of the first figure; also annex it to the third term of the 1st column, and multiply the result by this second figure, and add the product, after advancing it two places to the right, to the last term of the 2d column. Again, multiply this last result by this secondfigure of the root, and subtract the product from the dividend, and to the remainder annex the next period for a NEW DIVIDEND. Procesd with this second figure of the root precisely as was done with the first figure, and so continue, until all the periods have been brought down. NOrE.-This rule may be readily deduced from the rule for extracting the cube root of a polynomial, as given in my ALGEBRA. EXAMPLES. 1. Extract the cube root of 387420489. Operation. Number. Root. 1st COL. 2d COL. 387'420/489(729 7 49 343 14 147, 1st trial divisor. 44420, 1st div. 212 15124 30248 214 15552, 2d trial divisor. 14172489, 2d div. 2169 1574721 14172489 0 Explanation. The greatest cube of the first period, 387, is 343, whose root is 7, which we place to the right of the number for the first figure of the root sought. We also EXTRACTION OF THE CUBE ROOT. 195 place it for the first term of the first column; which, multiplied into itself, gives 7 x 7=49, for the first term of the 2d column, which, in turn, multiplied by 7, gives 49 x 7 343, which, subtracted from the first period, 387, leaves the remainder 44, to which, annexing the next period, 420, we get 44420 for the first dividend. Again, adding 7 to the first term, 7, of the 1st column, we get 7+7 = 1 4, for the second term of the 1st column, which, multiplied by 7, gives 14 x 7=98; this, added to the first term of the second column, gives 147 for the second term of the 2d column, or the first trial divisor. Again, adding 7 to the second term of the 1st column, we get 14 + 7 21, for the third term of the first column. The trial divisor, with two ciphers annexed, becomes 14700, which is contained 3 times in the first dividend, 44420. Since the trial divisor is less than the true divisor, it will sometimes give too large a quotient figure; such is the case in this present example, where 2 is the second figure of the root. This second figure, 2, of the root, annexed to the third term of the 1st column, gives 212; which, multiplied by 2, gives 424, which, being advanced two places to the right, must be added to 147, the last term of the 2d column. The sum 15124 will form the third term of the 2d column, which, multiplied by 2, gives 15124 x 2=30248, which, subtracted from the first dividend, leaves 14172 for the remainder, to which, annexing the next period, 489, we get 14172489 for the second dividend. Again, to the last term, 212, of the 1st column, adding 2, we get 214 for the next term; which, multiplied by 196 IIIGIIER ARlITHMETIC. 2, gives 428, which, added to 15124, gives 15552 for the second trial divisor. Again, adding 2 to 214, we get 216 for the fifth term of the 1st column. The second trial divisor, with two ciphers annexed, becomes 1555200, which is contained 9 times in the second dividend, 14172489; therefore, 9 is the third figure of the root, which, annexed to 216, gives 2169 for the last' term of the first column, which, multiplied by 9, gives 19521, which, advanced two places to the right, and then added to 15552, gives 1574721; this, multiplied by 9, gives 14172489, which, subtracted from the second dividend, leaves no remainder. 2. What is the cube root of 913517247483640899? Operalion. Number. Root. 1st COL. 2d COL. 913 51 7247'483'640'899(970299 9 81 729 18 243 184517 277 26239 183673 284 28227 844247483 29102 282328204 564656408 29104 282386412 279591075640 291069 28241260821 254171347389 291078 28243880523 25419728251899 2910879 2824414250211 25419728251899 0 3. What is the cube root of 10077696? Ans. 216. 4. What is the cube root of 2357947691? Ans. 1331. 5. What is the cube root of 42875? Ans. 35. EXTRACTION OF TIIE CUBE ROOT. 197 6. What is the cube root of 117649? Ans. 49. 7. What is the cube root of 350356403707485209? Ans. 704969. 8. What is the cube root of 75084686279296875? Ans. 421875. 9. What is the cube root of 7256313856? Ans. 1936. 10. What is the cube root of 106868920913284608? Ans. 474552. 11. The four following numbers, 2080913082956455142636, 4937801347Z510680732948, 7262810476410016163052, 214972108693241589340948, when added together, by taking two at a time, produce six distinct sums, each of which is a perfect cube. What are the six roots of these cube numbers? Ans. 19146344, 21062342, 60097344, Ans l 23021160, 60359866, 60571840. Case II. To extract the cube root of a decimal fraction, or of a number consisting partly of a whole number, and partly of a decimal value, we have this RULE. 1. Annex ciphers to the decimals, if necessary, so that the whole number of decimal places may be divisible By 3. II. Separate the decimals into periods of three figures each, counting from the decimal point towards the right, and proceed as in whole numbers. NOTE. —If the given number has not an exact root, there will be a remainder after all the periods have been brought down. The process may be continued by annexing ciphers for new periods. 17* 198 H IGHER A IT1'1l I fl' C EXAMPLES. 1. What is the cube root of 0'469640998917? Ans. 0'7773. 2. What is the cube root of 18'609625? Ans. 2'65. 3. What is the cube root of 1'25992105? Ans. 1'08005974, nearly. 4. What is the cube root of 2? Ans. 1] 25992105, nearly. 5. What is the cube root of 3? Ans. 1'442249, nearly. 6. What is the cube root of 1860867? Ans. 123. Case III. To extract the cube root of a vulgar fraction, or mixed number, we have this RULE. I. Reduce the fraction, or mixed number, to its simplest fractional form. II. Extract the cube root of the numerator and denominator separately, if they have exact roots; but when they have not, reduce the fraction to a decimal, and then extract the root by Case II. EXAMPLES. 1. What is the cube root of 2197? Ans. -7. 2. WVhat is the cube root of l8-7-3? Ans. 2 3 3. What is the cube root of 17-? Ans. 2'577, nearly. EXTRACTION OF THE CUBE ROOT. 199 4. What is the cube root of 5-? Ans. 1'726, nearly. 5. What is the cube root of 84? Ans. 0Y9353, nearly. 6. What is the cube root of 2? Ans. 0'8736, nearly.. 7. What is the cube root of 47?t Ans. 3'6173, nearly. 8- What is the cube root of 101-I? Ans. 4'6592, nearly. 9. What is the cube root of 93? Ans. 2'1085, nearly. Case IV. When there are many decimal places required in the root, we may, after obtaining one more decimal figure than half the required number, find the rest by dividing the remainder by the last term of the second column. Before dividing, we can omit from the right of the divisor so many figures as to leave but one more than the number of additional figures required in the root, observing to omit from the right of the dividend one figure less than was omitted in the divisor. The division must then be performed according to the abridged method, as explained under Art. 41. EXAMPLES. 1. What is tile cube root of 7, carried to 9 decimal places? 200 HIGHER ARITHMETIC. Operatiovn. Root. ISt UOL. 2d CoL. ( 191293118' I I 1 2 3 6 39 651 5859 48 1083 141 571 108871 108871 572 109443 32129 5732 10955764 21911528 5734 10967232 10217472 57369 1097239521 9875155689 57378 1097755923 342316311 573873 10977[7313919 329331941757 129841369243 10978 2006 1098 908 878 30 22 8 In this example, we proceed in the usual way, until we obtain 1'91293; the remainder is 12984369243; the last term of the second column is 109777313919; therefore, we obtain four more figures by dividing 12984369243 by 109777313919; but these four figures may be obtained with equal accuracy by dividing 12984 by 10977, which gives the remaining figures 1182. 2. Extract the cube root of ~=0025 to 13 decimal places. EXTRACTION OF THE CUBE ROOT. 201 Operation. Root. 1st COL. 2d COL. 0'250(0'6299605249474 6 36 216 12 108 34 182 11164 22328 184 11532 11672 1869 1170021 10530189 1878 1186923 1141811 18879 118862211 1069759899 18888 119032203 72051101 188976 11904354156 71426124936 188982 11905488048 624976064 18898805 1190549174974025 595274874870125 297011819129875 2381099 589019 476220 112799 107149 5650 4762 888 833 55 47 8 In this example, after obtaining seven decimal figures in the root, by the usual process, the remainder was 29701189129875, and the last term in the second column was 119054974974025; and, since we wish but six fig. ures by division, we reject seven figures from the right of the remainder, and eight figures from the right of the term of the second column, and then divide by the rule 202 IIT'11 ER A II1m'I'IC'. for abridging the work, Art. 41, and obtain the remaining figures of the root. 3. Extract the cube root of 9 to 9 decimals Operations. Root. 1st COL. 2d COL. 9(2'080083823 2 4 8 4 12 1 608 124864 998912 616 129792 108l 624008 12979' 6992064 1038375936512 496241063488 38939 10685 10383 302 259 43 39 4. What is the cube root of 152 to 5 decimal places? Arns. 2'50222. 5. What is the cube root of 3-.4 ~9265 to 8 decimals? Ans. 0'68278406. 6. What is the cube root of 0'0000031502374 to 13 decimals? Ans. 0'0146593403377. 7. What is the cube root of 2 to 21 decimals? Ans. 0'793700525984099737376. 8. What is the cube root of 4 to 10 decimals? Ans. 0'5227579585. 9. What is the cube root of 14 to 7 decimals? A /s. 0'9032157. APPLICATION OF THE CUBE ROOT. 203 10. What is the cube root of I 5 to 8 decimals? Ans. 0'13992727. E X A M P L E S INVOLVING THE PRINCIPLES OF THE CUBE ROOT. 79. IT is an established theorem of geometry, that all similar solids are to each other as the cubes of their like dimensions. 1. If a cannon ball 3 inches in diameter weigh 8 pounds, what will a ball of the same metal weigh, whose diameter is 4 inches? By the above theorem, we have 3": 43:: 8 pounds: 182 6 pound, for the answer. 2. Suppose the diameter of the sun to be 887681 miles; the diameter of the earth, 7912 miles. How many times greater in bulk is the sun than the earth? (887681) 3 = 699472706450842241; (7912)3 — =495289174528; 699472706450842241' 495289174528 -- 1412251 times, nearly. 3. How many cubic quarter inches can be made out of a cubic inch? Ans. 64. 4. Required the dimensions of a rectangular box, which shall contain 20000 solid inches; the length, breadth, and depth being to each other as 4, 3, and 2. 204 HIGHER AR'IIMETIC. 2 X 1 X XI- = 2 o ~, whose cube root is 5 20_ 9'4103, nearly. [ 9'4103 x 4- 37'6412, length. Ans.j 9'4103 x 3 28'2309, breadth..9'4103 x 2- 18'8206, depth. Or, as follows: If we were to augment the width of this box, so as to make it as wide as it is long, its volume would become ~ of 20000 -266669-. Again, if we augment the depth of this new box, so that it may be as deep as it is wide, and as it is long, its volume will become 2 times 266663 =533333, which is the contents of a cubical box, whose side is equal to the length of the original box. Hence, 53333= 37'641, nearly, for the length. The width is 3 of this length, and the depth is I this length. 5. What is the side of a cube which will contain as much as a chest 8 feet 3 inches long, 3 feet wide, and 2 feet 7 inches deep? Ans. 47'9843 inches. 6. Four ladies purchased a ball of exceeding fine thread, 3 inches in diameter. What portion of the diameter must each wind off so as to share of the thread equally? Solution. After the first one had wound off her share, the ball which remained would contain 3 as much thread as it did in the first place. Therefore, its diameter was 3 3 - - 2'72568 inches, nearly. The diameter, after the second one had wound off her share, was 3 -2 = 23 4-2'38110 inches, nearly. ROOTS OF ALL POWERS. 205 The diameter, after the third one had wound off her share, was 3 4- =3 /2_= 1'88988 inches, nearly. Hence, the portions of the diameter which they must wind off are as follows: Inches, nearly. The 1st lady must wind off 3'00000-2'72568 = 0'274352 2d " " " 2' 72568-2'38110=0 34458 3d " " " "2'38110-1 88988=0'49122 4th " " " 1'88988 =1'88988 ROOTS OF ALL POWERS. 0S. WHENEVER the index denoting the root required is a composite number, the root can be found by successive extractions of the roots denoted by the prime factors of the-original index. Thus, the 4th root may be found by extracting the-2d root twice in succession. The 6th root may be obtained by extracting the 3d root of the second root. The 8th root may be found by extracting the 2d root three times in succession. When the index denoting the root is a prime,.we must have some direct method of obtaining the root. By a similar train of reasoning, as was used in deducing the rule for the cube root, we determine, in general, for any root, the following RULE. 1. Point the number off into periods of as many figures each as there are units in the index denoting the root. 18 206 HIGHIER ARITHMIETIC. II. Find, by trial, theJigure of the first period, which will be the first figure of the root; place this figure to the left, in a column called the FIRST COLUMN. Then multiply it by itself, and place the product for the first term of the SECOND- COLUMN. This, multiplied by the same figure, will give the first term of the THIRD COLUMN. Thus continue until the number of columns is one less than the units in the index denoting the root. Multiply the term in the LAST COLUMN by the same figure, and subtract the product from the first period, and to the remainder bring down the next period, and it will form the FIRST DIVIDEND. Again, add this same figure to the term of the FIRST COLUMN, multiply the sum by the same figure, and add the product to the term of the SECOND COLUMN; which, in turn, must be multiplied by the same figure, and added to the term of the TIIIRD COLUMN, and so on, till we reach the LAST COLUMN, the term of which willform the FIRST TRIAL DIVISOR. Again, beginning with the FIRST COLUMN, repeat the above process until we reach the column next to the last. And so continue to do, until we obtain as many terms in the FIRST COLUMN as there are units in the index denoting the root; observing, in each successive operation, to terminate on the column of the next inferior order. III. Seek how many times the FIRST TRIAL DIVISOR, when there are annexed to it as many ciphers, less one, as there are units in the index, is contained in the FIRST DIVIDEND; the quotient figure will be the second figure of the root. Then proceed with this figure the same as ROOTS OF ALL POWERS. 207 was done with the first igure; observing to advance the terms of the different columns as many places to the right as the number expressing the order of the column; that is, advancing the terms of the FIRST COLUMN one place, those of the SECOND COLUMN two places, and so for the succeeding columns. After completing the requisite number of terms in the different columns, by means of this second figure of the root, then proceed to obtain the third figure of the root in the same way as the second figure was obtained; and in this way the operation can be continiued until all the periods are brought down. If there is still a remainder, the process can be extended by forming periods of ciphers. EXAMPLES. 1. What is the fifth root of 36936242722357? Operation. Ist CoL. 2d COL. 3d COL. 4th CoL. Root. 5 25 125 625 3693'62427/22357(517 10 75 500 3125 3125 15 150 1250 32525251 56862427 20 250 1275251 33826005 32525251 251 25251 1300754 347673946051 2433717622357 252 25503 1326510 2433717622357 253 25756 1344842293 0 254 26010 2557 2618899 2. What is the 7th root of 1231171548132409344? Operation. Root. 1st CoL. 2d CoL. 3d COL. 4th COL. 5th COL. 6th COL. 12311/7154813/2409344(384 3 9 27 81 243 729 2187 6 27 108 405 1458 5103 101247154813 9 54 270 1215 5103 11568197824 92545582592 12 90 540 2835 808149728 21076554688 87015722212409344 15 135 945 37231216 1188544608 21753930553102336 87015722212409344 18 189 1110152 47549360 1663938528 218 20644 1289768 59424240 169343966275584 226 22452 1484360 72979760 234 24324 1694440 737528368896 242 26260 1920520 250 28260 1932692224 258 30324 2664 3043056 3. What is the eleventh root of 11? [The work in this question is so lengthy, we have been compelled to make use of two pages in our operation.] Operation. ]st COL. 2d COL. 3d COL. 4th COL. 5th COL. 6th COL. 7th COL. 1 1 1 1 1 1 1 2 3 4 5 6 7 8 3 6 10 15 21 28 36 4 10 20 35 56 84 1.20 5 15 35 70 126 210 330 6 21 56 126 252 462 4438023168 ~ 7 28 84 210 462 569011584 5822306304 8 36 120 330 53505792 692141568 7488297728 9 45 165 3652896 61564992 832995712 9474862720 10 55 176448 4029600 70427072 993282496 11824496640 112 5724 188352 4431040 80143392 1174816960 123951454696751104 * 114 5952 200720 4858160 90767232 1379524608 129850860737626112 116 6184 213560 5311920 102353824 1426622074187776 118 6420 226880 5793296 114960384 1474851510218752 120 6660 240688 6303280 11774366546944 122 6904 254992 6842880 12057359007744 124 7152 269800 69582036736 126 7404 285120 70748115200 128 7660 288309184 130 7920 291519616 1324 797296 1328 802608 8th COL. 9th COL. 10h COL.? 1 1 1 t k 9 10 11 s 4 45 55 321504185344 165 105752092672 681091006464'' 25376046336 179793410560 8067509802924592070656 37020658944 283787919360 94513868057140984283136 CD 51997254400 314149934571148017664 C 70946979840 346589563554098053120 D r 7590503802787004416 L: 8109907245737508864 $o ~, 11(1'2436, nearly, which is a trifle too great. Zs 10 643008370688 j -356991629312 o ~ 32270039211698368282624 ~:) v4 34291123719501631717376 2836 co XD 593 ~ 567 26 ARITHIMETICAL PROGRESSION. 211 CHAPTER X. ARITHMETICAL PROGRESSION. 81i. A SERIES of numbers which succeed each other regularly by a common difference, is said to be in arithmetical progression. When the terms are constantly increasing, the series is an arithmetical progression ascending. When the terms are constantly decreasing, the series is an arithmetical progression descending. Thus, 1, 3, 5, 7, 9, &c., is an ascending arithmetical progression; and 10, 8, 6, 4, 2, is a descending arithmetical progression. In arithmetical progression, there are five things to be considered: 1. The first term. 2. The last term. 3. The common difference. 4. The number of terms. 5. The sum of all the terms. These quantities are so related to each other, that any three of them being given, the remaining two can be found. 212 HIGhtER ARITIIMETICl If we denote the five things by the numerals 1, 2, 3, 4, 5, they may be taken by threes, as follows: 1, 2, 3, giving 4 and 5, making 2 cases. 1, 2,4 " 3 " 5 " 2 " 1, 2, 5 " 3 " 4 " 2 " 1, 3, 4 " 2' 5 " 2 2 1, 3,5 " 2 " 4 " 2 " 1,4, 5 " 2 " 3 " 2 " 2, 3,4 " 1 " 5 2 " 2, 3, 5 " 1 " 4 " 2 " 2, 4, 5 " 1 " 3 " 2 " 3, 4, 5 " 1 " 2 " 2 " Hence, there must be 20 distinct cases arising from the different combinations of these five quantities. To give a demonstration to all the rules of these 20 cases, would be a very difficult task for the ordinary processes of arithmetic; we will, therefore, content ourselves with demonstrating a few of the most important of them. Case L By our definition of an ascending arithmetical progression, it follows that the second term is equal to the first, increased by the common difference; the third is equal to the first, increased by twice the common difference; the fourth is equal to the first, increased by three times the common difference; and so on, for the succeeding terms. Hence, when we have given the first term, the common difference, and the number of terms, to find the last term, we have this ARITHMETICAL PROGRESSION. 213 RULE. To the first term add the product of the common difference into the number of terms, less one. EXAMPLES. 1. What is the 100th term of an arithmetical progres sion, whose first term is 2, and common difference 3? In this example, the number of terms, less one, is 99; which, multiplied by the common difference, 3, gives 297, which, added to the first term, 2, makes 299 for the 100th term. 2. What is the 50th term of the arithmetical progression, whose first term is 1, the common difference being -? Ans. 251. 3. A man buys 10 sheep, giving $1 for the first, $3 for the second, $5 for the third, and so increasing in arithmetical progression. What will the last sheep cost at that rate? Ans. $19. 4. A person bought 100 yards of cloth; he gave 2 s. 6 d. for the first yard, 2 s. 10 d. for the second yard, 3 s. 2d. for the third yard, and so continued to give 4d. more for each yard than he gave for the preceding one., How much did he give for the last yard? Ans. ~1 15s. 6d. Case II. From the nature of an arithmetical progression, we see that the second term added to the next to the last term, is equal to the first added to the last, since the second term is as much greater than the first, as the next to the last is less than the last. After the same 214 HIGHER ARITHMETIC. method of reasoning, we infer that the sum of any two terms equidistant from the extremes is equal to the sum of the extremes. Hence, it follows that the terms will average just half the sum of the extremes. Therefore, when we have given the first term, the last term, and the number of terms, to find the sum of all the terms, we have this RULE. Multiply half the sum of the extremes by the number of terms. EXAMPLES. 1. The first term of an arithmetical progression is 2, the last term is 50, and the number of the terms is 17. What is the sum of all the terms? In this example, half the sum of the extremes is 2 ~ 50 2+5 —26; this, multiplied by the number of terms, gives 26 x 17=442, for the sum required. 2. The first term of an arithmetical progression is 13, the last term is 1003, and the number of terms is 100. What is the sum of the progression? Ans. 50800. 3. A person travels 25 days, going II miles the first day, and 135 the last day; the miles which he traveled in the successive days form an arithmetical progression. How far did he go in the 25 days? Ans. 1825 miles. 4. Bought 7 books, the prices of which are in arithmetical progression. The price of the first was 8 shillings, and the price of the last was 28 shillings. What did they all come to? Ans. ~6 6 s. ARITHMETICAL PROGRESSION. 215 Case III. By Case I. we see that the last term is equal to the first term, increased by the product of the common difference into the number of terms, less one. Hence, the first term must equal the last term, diminished by the product of the common difference into the number of terms, less one. Therefore, when we have given the last term, the number of terms, and the common difference, to find the last term, we have this RULE. Front the last term subtract the product of the common difference into the number of terms, less one. EXAMPLES. 1. The last term of an arithmetical progression is 375, the common difference is 7, and the number of terms is 54. What is the first term? In this example, the common difference, multiplied by the number of terms, less one, is 7x53=371, which, subtracted from the last term, gives 375-371=4, for the first term. 2. The last term of an arithmetical progression is 393, the common difference is 2, and the number of terms is 59. What is the first term? Ans. 23. 3. A note is paid in 15 annual instalments; the payments are in arithmetical progression, whose common difference is 3; the last payment is 49 dollars. What is the first payment? Ans. $7. 216 HIGHER ARITHMETIC. Case IV. From Case I. we see that the product of the common difference into the number of terms, less one, is equal to the last term diminished by the first. Therefore, the difference of the last and first terms, divided by the common difference, is equal to the number of terms, less one. Hence, when we have given the first term, the last term, and the common difference, to find the number of terms, we have this RULE. Divide the difference of the extremes by the common difference, and to the quotient add one. E X AMP L ES. 1. The first term of an arithmetical progression is 5, the last term 176, and the common difference is 3. What is the number of terms? In this example, the difference of the extremes is 176 —5= —171; this, divided by the common difference, gives 7- =57; which, increased by 1, becomes 58, for the number of terms required. 2. The first term of an arithmetical progression is 11, the last term is 88, and the common difference is 7. What is the number of terms? Ans. 12. 3. A note becomes due in annual instalments, which are in arithmetical progression, whose common difference is 3; the first payment is 7 dollars, the last payment is 49 dollars. What is the number of instalments? Ans. 15. r, lP I 1t1 XETI CAL P ROG RES SION. 217 CJase V. We learn from Case I. that the product of the corn mon difference into the number of terms, less one, is equal to the last term diminished by the first. Therefore, the difference of the last and first terms, divided by the number of terms, less one, will give the common difference. Hence, when we have given the first term, the last term, and the number of terms, to find the common difference, we have this RULE. Divide the difference of the ext-emnes by the number of terms, less one. EXAMPLES. 1. The first term of anl arithmetical progression is 5, the last term is 176, and the number of terms 58. What is the common difference? In this example, the difference of the extremes is 171; which, divided by the number of terms, less one, becomes — 71 -3, for the common difference. 2. A person performs a journey in 17 days; the distances traveled on the successive days were in arithmetical progression; the first day he went 4 miles, and the last day he went 84. How many miles more did he go on each day, than on the preceding day? Ans. 5 miles. 3. A man has 7 sons, whose ages are in arithmetical progression; the age of the eldest is 41 years, the youngest is 5 years old. How many years is the common difference of their ages? Ans. 6 years. 19 218 HIGHER ARITHMETIC. Case VI. By Case II. we know that the sum of all the terms of an arithmetical progression is equal to half the sum of the extremes multiplied into the number of terms; therefore, the number of terms is equal to the sum of all the terms divided by half the sum of the extremes. Hence, when we have given the first term, the last term, and the sum of all the terms, to find the number of terms, we have this RULE. Divide the sum of all the terms by half the sum of the extremes. EXAMPLES. 1. The first term of an arithmetical progression is 1, the last term is 1001, and the sum of all the terms is 251001. What is the number of terms? In this example, half the sum of the extremes is 1001+ ~1 =501; then, dividing the sum of all the terms by this, we obtain 252 -0o0 =501, for the number of terms. 2. In a triangular field of corn, the number of hills in the successive rows are in arithmetical progression; in the first row there is but one hill, in the last row there are 81 hills, and the whole number of hills in the field is 1681. How many rows are there? Ans. 41. 3. A man bought a certain number of yards of cloth, for $152'50, giving 4 cents for the first yard, and in creasing regularly on each succeeding yard, up to the ARITHMETICAL PROGRESSION. 219 last yard, for which he gave $3-01. How many yards of cloth did he purchase? Ans. 100. Case VII. We also infer from Case II. that the sum of all the terms, divided by half the number of terms, will give the sum of the extremes. Therefore, if from the quotient of the sum of all the terms, divided by half the number of terms, we subtract the last term, we shall have left the first term. Hence, when we have given the last term, the numbei of terms, and the sum of all the terms, to find the first term, we have this RULE. From the quotient of the sum of all the terms, divided by half the number of terms, subtract the last term. EXAMPI, ES. 1. If the last term of an arithmetical progression is 170, the number of terms 50, and the sum of all the terms 4450, what is the first term? In this example, the sum of all the terms, divided by half the number of terms, is 44A 0 178; from which, subtracting the last term, we obtain 178 —170-8, for the first term. 2. A person wishes to discharge a debt of $1125 in 18 annual payments, which shall be in arithmetical progression. How much must his first payment be, so as to bring his last payment $120? Ans. $5. 3. The miles which a person travels in 19 successive days, form an arithmetical progression, whose last term 220 HIGHER ARITHMETIC. is 80, the sum of all the terms 950. How many miles does he travel the first day? Ans. 20 miles. Case VIII. From what has been said under Case VII. we infer that the first term subtracted from the quotient of the sum of all the terms divided by half the number of terms, will give the last term. Hence, when we have given the sum of all the terms, the first term, and the number of terms, to find the last term, we have this RULE. From the quotient of the sum of all the terms, divided by half the number of terms, subtract the first term. EXAMPLES. 1. If the first term of an arithmetical progression is 7, the number of terms 1000, and the sum of all the terms 560000, what is the last term? In this example, the sum of all the terms, divided by half the number of terms, gives 5- o0o =1120; from which subtract the first term, we get 1120 —7=1113, for the last term. 2. If the first term of an arithmetical progression is 7, the number of terms 16, and the sum of all the terms 142, what is the last term? Ans. 10.03 3. The first term of an arithmetical progression is 13, the number of terms 100, and the sum of all the terms 50300. What is the last term? Arns. 993. NOTE. —The remaining cases are obtained by combining the conditions of Cases I. and II. ARIT HMETICAL PROGRESSION. 221 Case XI. Given the common difference, the number of terms, and the sum of all the terms, to find the first term. RUILE. Divide the sum of the terms by the number of terms; from this quotient subtract half the product of the common difference into the number of terms, less one. EXAMPLES. 1. The common difference of the terms of an arithrnetical prcgression is 7, the number of terms 54, and the sum of all the terms is 10233. What is the first term? In this example, the sum of the terms, divided by the number of terms, is 1891-; half the product of the number of terms, less one, into the common difference, is 185k-; which, subtracted from 189-, leaves 4 for the first term. 2. The common difference of the terms of an arithinetical progression is 2, the number of terms is 59, and the sum of all the terms is 1180. What is the first term? Ans. 2. 3. A father divides $2000 among five sons, so that each should receive $40 more than his next younger brother. What is the share of the youngest? Ans. $320. Case X. Given the common difference, the number of terms, and the sum of all the terms, to find the last term. 19' 222 HIIIGIfER ARITIIMETIC. RULE. To the quotient of the sum of the terms, divided by the number of terms, add half the product of the common difference into the number of terms, less one. EXAMPLES. i. The common difference of the terms of an arithmetical progression is 6, the number of terms is 7, and the sum of all the terms is 161. What is the last term? In this example, the sum of the terms, divided by the number of terms, is -7 1=23. Again, the common difference multiplied into the number of terms, less one, is 6 x 6=36, the half of which is 18, which, added to 23, gives 41 for the last term. 2. The common difference of the terms of an arithmetical progression is 7, the number of terms is 54, and the sum of all the terms is 10233. What is the last term? Ans. 375. 3. The common difference of the terms of an arithmetical progression is 6, the number of terms is 14, and the sum of all the terms is 4970. What is the last term? Ans. 394. Case XI. Given the first term, the common difference, and the number of terms, to find the sum of all the terms. RULE. To twice the first term, add the product of the common difference into the number of terms, less one; multiply this sum by half the number of terms. ARITHlMETICAL PROGRESSION. 223 EXAMPLES. 1. The first term of an arithmetical progression is 37, the common difference is 11, and the number of terms 99. What is the sum of all the terms? In this example, the product of the common difference into the number of terms, less one, is 11 x 98- 1078; this, added to twice the first term, gives 74+1078= 1152, which, multiplied by half the number of terms, gives 57024 for the sum of all the terms. 2. The first term of an arithmetical progression is 7, the common difference is 1, and the number of terms 37. What is the sum of all the terms? Ans. 1258. 3. A person buys 37 sheep, paying for them in arithmetical progression; for the first he gives 3 shillings, and increases I shilling for each succeeding one. How much did they all come to? Ans. ~38 17s. Case XIL ~Given the first term, the common difference, and the last term, to find the sum of all the terms. RUL E. Divide the difference of the squares of the last and first terms by twice the common difference, and to this quotient add half the sum of the last and first terms. EXAiMPT, ES. 1. The first terill of an arithmetical progression is 16, the common difference is 2, and the last term 100. What is the sum of all the terms? 224 HIIGIIERI ARITHMETIC. In this example, the difference of the squares of the last and first terms is 9744, which, divided by twice the common difference, gives 2436; this, increased by half the sum of the last and first term, becomes 2494, for the sum of all the terms. 2. The first term of an arithmetical progression is 5, the common difference is 7, and the last term is 75. What is the sum of all the terms? Ans. 440. 3. The first term of an arithmetical progression is 8, the common difference 3, and the last term 170. What is the sum of all the terms? Ans. 4895. C(ase XIII. Given the common difference, the number of terms, and the last term, to find the sum of all the terms. RU, E. From twice the last term, subtratct the product of the common difference into the number of terms, less one, multiply this revmainder by half the number of terms. EXAMPLES. 1. The commnit-on difference of the terms of an arithmetical progression is 11, the number of terms is 19, and the last term is 199. What is the sum of all the terms? In this example, the product of the common difference into the number of terms, less one, is 11 x 18 - 198; this, subtracted from twice the last term, gives 200, which, multiplied by half the number of terms, becomes 1900, for the sum of all the terms. AR1TIRMETICAL PROGRESSION. 225 2. The common difference of the terms of an arithmetical progression is 15, the number of terms is 47, and the last term is 545. What is the sum of all the terms? Ans. 9400. 3. The common difference of the terms of an arithmetical progression is 4-, the number of terms is 100, and the last term is 1000. What is the sum of all the terms? Ans. 77725. C(ase XIV. Given the first term, the number of terms, and the sum of all the terms, to find the common difference. R U LE. From twice the sum of the terms, subtract twice the product of the first term into the number of terms; divide this remainder by the product of the number of terms into the number of terms, less one. EXAMPLE S. 1. The first term of an arithmetical progression is 21, the number of terms is 50, and the sum of all the terms is 3500. What is the common difference? In this example, twice the product of the first term into the number of terms, is 2100; which, subtracted from twice the sum of the terms, gives 4900; the number of terms multiplied into the number of terms, less one, is 2450; hence, 4900, divided by 2450, gives 2 for the common difference. 2. The first term of an arithmetical progression is 7, the number of terms is 13, and the sum of all the terms is 1394. What is the common difference? Ans. 1-. 226 I I G II E IR A R I TIE l TIC. 3. The first term of an arithmetical progression is', the number of terms is 26, and the sum of all the terms is 60-. What is the common difference? Ans. - Case X V. Given the first term, the last term, and the sum of all the terms, to find the common difference. RULE. Divide the difference of the squares of the last and first term by twice the sum of all the terms, diminished by the sum of the last and Jirst term. E X AMPLE 1 E S. 1. The first term of an arithmetical progression is -, the last term is 204-, and tile sum of all the terms is 139k. What is the common difference? In this example, the difference of the squares of the last and first term is (204)2 —(-)2=428-; twice the sum of all the terms, diminished by the sum of the last and first term, is 2574. Dividing, 4284 by 2574, we get A-12, for the common difference. 2. The first term of an arithmetical progression is 8, the last term is 170, and the sum of all the terms is 4895. What is the common difference? Ans. 3. 3. The first term of an arithmetical progression is 12, the last term is 102, and the sum of all the terms is 912. What is the common difference? Ans. 6. Case XVI. Given the number of terms, the last term, and the sum of all the terms, to find the common difference. ARITHMETICAL PROGRESSION. 227 RU LE. From twice the product of the number of terms into the last term, subtract twice the sum of all the terms; divide the remainder by the product of the number of terms into the number of terms, less one. EXAMPLES, 1. The number of terms of an arithmetical progression is 17, the last term is 50, and the sum of all the terms is 442. What is the common difference? In this example, twice the product of the number of terms into the last term is 2 x 17 x 50 — 1700; the product of the number of terms into the number of terms, less one, is 17x 16-272; also 1700, diminished by twice the sum of all the terms, becomes 816, which, divided by 272, gives 3 for the common difference. 2. The number of terms of an arithmetical progression is 14, the last term is 14, and the sum of all the terms is 105. What is the common difference? Ans. 1. 3. The number of terms of an arithmetical progression is 7, the last term is 41, and the sum of all the terms is 161. What is the common difference? Ans. 6. NOTE.-The remaining four cases require for their solution the Extraction of the Square Root. Case XVII. Given the first term, the common difference, and the sum of all the terms, to find the number of terms. 228 HIGHER ARITIHMETIC RUI E. Subtract the colmmon diffperence from twice the first term; divide the remnainder by twice the common difference; to the square of this quot'ient add the quotient of twice the sum of all the teerms divided by the common difference; extract the sqtuare root of the snam; then divide twice the first term, dimninished by the common difference, by twice the common difference, and subtract this quotient from the root just found. EXAMPI, ES. 1. The first term of an arithmetical progression is 7, the common difference is I, and the sum of all the terms is 142. What is the number of terms? In this example, the common difference, subtracted from twice the first terrm, gives 133, which, divided by twice the common difference, gives 27T, which, squared, becomes 756~. Twice the sum of all the terms, divided by the common difference, gives 1136, which, added to 756k, gives 1892~, the square root of which is 43;i from this, subtracting 27', we get 16 for the number of terms. 2. The first term of an arithmetical progression is 2, the common difference is 3, and the sum of all the terms is 442. What is the number of terms? Ans. 17. 3. The first term of an arithmetical progression is 4, the common difference is -, and the sum of all the terms is 60-. What is the number of terms? Ans. 26. Case XVIII. Given the common difference, the last term, and the sum of all the terms, to find the number of terms. AlR I'HME'TICAL I PROGRESSION. 229 RU LE. To twice the last term add the common difference; divide the sum by twice the common difference; square the quotient, and from this square subtract the quotient of twice the sum of the terms divided by the common difference; extract the square root of the remainder; then subtract this root from the quotient of the sum of twice the last term and common difference, divided by twice the common difference. EXAMPLES. 1. The common difference of the terms of an arithmetical progression is 3, the last term is 35k, and the sum of all the terms is 1900. What is the number of terms? In this example, twice the last term, increased by the common difference, is 71, which, divided by twice the common difference, gives 107; this, squared, becomes 11449. Again, twice the sum of all the terms, divided by the common difference, gives 11400; this, subtracted from 11449, gives 49, whose square root is 7. Subtracting this root from 107, we get 100 for the number of terms. 2. The common difference of the terms -of an arithmetical progression is 8, the last term is 37, and the sum of all the terms is 60-. What is the number of terms? Ans. 26. 3. The common difference of the terms of afi arithmetical progression is 1, the last term is 14, and the sum of all the terms is 105. What is the number of terms? Ans. 14. 20 230 IlIGHIER ARITIIMETIC. Case XIX. Given-the first term, the common difference, and the sum of all the terms, to find the last term. R U L E. From the first termst subtract half the -commzon difference, and to the square of the remainder add twice the product of the common difference into the sum of all the terms; then extract the square root; which, diminished by half the common difference, will give the last term. EXAMP, ES. 1. The first term of an arithmetical progression is 4, the common difference is 7, and the sum of all the terms is 10233. What is the last term? In this example, half the common difference, subtracted from the first term, gives I, which, squared, is {; this, added to twice the product of the common difference into the sum of all the terms, which is 2 x 7x 10233143262, gives 573A-04 9, whose square root is 7-_7_; from this, subtract half the common difference, and we find 757 7 - 7. 0 = 375 for the last term. 2. The first term of an arithmetical progression is 23, the common difference is -, and the sum of all the terms is 1180. What is the last term? Ans. 391. 3. A man has several sons, whose ages are in arithmetical progression; the age of the youngest is 5 years, the common difference of their ages is 6 years, and the sum of all their ages is 161. What is the age of the eldest? Ans. 41 years. ARITHMETICAL PROGRESSION. 231 Case XX. Given the common difference, the last term, and the sum of all the terms, to find the first term. RULE. Add half the common difference to the last term, and from the square of the sum subtract twice the product of the common difference into the sum of all the terms; then extract the square root of the remainder, and to this root add half the common difference. EXAMPLES. 1. The common difference of the terms of an arithmetical progression is 4, the last term is 1008, and the sum of all the terms is 127512. What is the first term? In this example, half the common difference, added to the last term, gives 1010, which, squared, is 1020100; twice the product of the common difference into the sum of all the terms is 1020096, which, subtracted from 1020100, leaves 4, the square root of which is 2; this, increased by half the common difference, becomes 4 for the first term. 2. The common difference of the terms of an arithmetical progression is 3, the last term is 49, and the sum of all the terms is 420. What is the first term? Ans. 7. 3. The common difference of the terms of an arithmetical progression is 10, the last term is 1003, and the sum of all the terms is 50800. What is the first term? Ans. 13. 232 HIGHER ARITIIMETIC. CHAPTER XI. GEOMETRICAL PROGRESSION. 3~2. A SERIES of numbers which succeed each other regularly, by a constant multiplier, is called a geometrical.progression. This constant factor, by which the successive terms are multiplied, is called the ratio. When the ratio is greater than a unit, the series is called an ascending geometrical progression. When the ratio is less than a unit, the series is called a descending geometrical progression. Thus, 1, 3, 9, 27, 81, &c., is an ascending geometrical progression, whose ratio is 3. And 1, 4-, 9, -, &c., is a descending geometrical progression, whose ratio is 4. In geometrical progression, as in arithmetical progression, there are five things to be considered: 1. The first term. 2. The last term. 3. The common ratio. 4. The number of terms. 5. The sum of all the terms. These quantities are so related to each other, that any three being given, the remaining two can be found. GEOMETRICAL, IPROGRESSION. 233 Hence, as in arithmetical progression, it may be shown that there must be 20 distinct cases arising from the different combinations of these five quantities. The solution of some of these cases requires a knowledge of higher principles of mathematics than can be detailed by arithmetic alone. We will give a demonstration of the rules of some of the most important cases. Case 1'. By the definition of a geometrical progression, it follows that the second term is equal to the first term multiplied by the ratio; the third term is equal to the first term, multiplied by the second power of the ratio; the fourth term is equal to the first term, multiplied by the third power of the ratio; and so on, for the succeeding terms. Hence, when we have given the first term, the ratio, and the number of terms, to find the last term, we have this RULE. Multiply the first term by the power of the ratio whose exponent is one less than the number of terms. EXAM P L ES. 1. The first term of a geometrical progression is 1, the ratio is 2, and the number of terms is 7. What is the last term? In this example, the power of the ratio, whose exponent is one less than the number of terms, is 26=64, which, multiplied by the first term, 1, still remains 64, for the last term. 20* 234 HIGHIER ARITHMETIC. 2. The first term of a geometrical progression is 5, the ratio is 4, and the number of terms 9. What is the last term? Ans. 327680. 3. A person traveling, goes 5 miles the first day, 10 miles the second day, 20 miles the third day, and so on, increasing in geometrical progression. If he continue to travel in this way for 7 days, how far will he go the last day? Ans. 320 miles. Case 11. If we multiply all the terms of a geometrical progression by the ratio, we shall obtain a new progression, whose first term equals the second term of the old progression; the second term of our new progression will equal the third term of the old progression, and so on for the succeeding terms. Hence, the sum of the old progression, omitting the first term, equals the sum of the new progression, omitting its last term. The sum of the new progression is equal to the old progression repeated as many times as there are units in the ratio. Therefore, the difference between the new progression and the old progression is equal to the old progression repeated as many times as there are units in the ratio, less one. But we also know tehat the difference between these progressions is equal to the last term of the new progression diminished by the first term of the old progression; and, since the new pro-gression was formed by multiplying the respective terms of thec old progression by the ratio, it follows tllat the last termry of the nvew progression is equal to the last term of the old progressico repeated as many times as there are units in the ratio. GEOMIETRICAL PROGRESSION. 235 Therefore, the last term of the new progression, diminished by the first term of the old progression, is equal to the last term of the old progression repeated as many times as there are units in the ratio and diminished by the first term of the old progression. Hence, we finally obtain this condition: That the sum of all the terms of a geometrical progression, repeated as many times as there are units in the ratio, less one, is equal to the last term multiplied by the ratio and diminished by the first term. Hence, when we have given the first term of a geometrical progression, the last term, and the ratio, to find the sum of all the terms, we have this RULE. Subtract the first term from the product of the last term into the ratio; divide the remainder by the ratio, less one. EXAMPLES. 1. The first term of a geometrical progression is 4, the last term is 78732, and the ratio is 3. What is the sum of all the terms? In this example, the first term, subtracted from the product of the last term into the ratio, is 236192, which, divided by the ratio, less one, gives 118096, for the sum of all the terms. 2. The first term of a geometrical progression is 5, the last term is 327680, and the ratio is 4. What is the sum of all the terms? Ans. 436905. 3. A person sowed a peck of wheat, and used the 236 HIGHER ARITHMETIC. whole crop for seed the following year; the produce of this second year again for seed the third year, and so on. If, in the last year, his crop is 1048576 pecks, how many pecks did he raise in all, allowing the increase to have been in a four-fold ratio? Ans. 139S101 pecks. Case I.II. Since by Case I. the last term is equal to the first term multiplied into a power of the ratio whose exponent is equal to the number of terms, less one, it follows that the first term is equal to the last term divided by the power of the ratio whose exponent is one less than the number of terms. Hence, when we have given the last term, the ratio, and the number of terms, to find the first term, we have this RULE. Divide the last term by a power of the ratio whose exponent is one less than the number of terms. EXAMPLES. 1. The last term of a geometrical progression is 1048576, the ratio is 4, and the number of terms is 11. What is the first term? In this example, the ratio, 4, raised to a power whose index is 10, one less than the number of terms, is 4'1 = 1048576;.'. 1048576, divided by 1048576, gives 1 for the first term. 2. A man has 6 sons, among whomn he divides his estate in a geometrical progression, whose ratio is 2; GEOMETRICAL PROGRESSION. 237 the last son received $4800. How much did the first son receive? Ans $150. 3. A person bought 10 bushels of wheat, paying for it in geometrical progression, whose ratio is 3; the last bushel cost him $196'83. What did he give for the first bushel? Ans. 1 cent. Case IV. We also discover from Case I. that the last term divided by the first term, will give the power of the ratio, whose exponent is the number of terms, less one. Hence, when we have given the first term, the last term, and the number of terms, to find the ratio, we have this RULE. Divide the last term by the first term; extract that root of the quotient which is denoted by the number of terms, less one. EXAMPLES. 1. The first term of a geometrical progression is 1, the last term is 64, and the number of terms is 7. What is the ratio? In this example, the last term, divided by the first term, is 64; the number of terms, less one, is 6,.. we must extract the 6th root of 64; we first extract the square root, which is 8, we now extract the cube root of 8, which is 2, for the ratio. 2. In a country, during peace, the population increased every year in the same ratio, and so fast that in the space of 5 years it became from 10000 to 14641 souls. By what ratio was the increase, yearly? Ans. -I. 238 HIG ER ARIT IME'IlC. 3. The first term of a geometrical progression is 4, the last term is 78732, and the number of terms is 10. What is the ratio? Ans. 3. Case V. If in Case II. we write the product of the first term into the power of the ratio, whose exponent is the number of terms, less one, instead of the last term, as drawn from Case I., we shall have the sum of all the terms, repeated as many times as there are units in the number of terms, less one, equal to the power of the ratio, whose exponent is equal to the number of terms diminished by one, and multiplied by the first term. Hence, when we have given the first term, the ratio, and the number of terms, to find the sum of all the terms, we have this RULE. From the power of the ratio, whose exponent is the number of terms, subtract one, divide the remainder by the ratio, less one, and multiply the quotient by the first term. EXAMPLES. 1. The first term of a geometrical progression is 3, the ratio is 4, and the number of terms is 9. What is the sum of all the terms? In this example, the ratio, raised to a power whose exponent is the number of terms, is 49 = 262144; this, diminished by one, becomes 262143, which, divided by 3, gives 87381; this, multiplied by the first term, becomes 87381 x 3-262143, for the sum of all the terms. GEOMETRICAL PROGRESSION. 239 2. A king in India, named SHERAN, wished (according to the Arabic author ASEPIIAD,) that SESSA, the inventor of chess, should himself choose a reward. He requested the grains of wheat which arise when 1 is calculated for the first square of the board, 2 for the second square, 4 for the third, and so on; reckoning for each of the 64 squares of the board twice as many grains as for the preceding. When it was calculated, to the astonishment of the king, it was found to be an enormous number. What was it? Ans. 18446744073709551615 grains. 3. A gentleman married his daughter on New-Year's day, and gave her husband 1 shilling towards her portion, and was to double it on the first day of every month during the year. What was her portion? Ans. ~204 15s. Case VI. We know from Case V. that the sum of all the terms multiplied by the ratio, less one, is equal to one subtracted from the power of the ratio, whose exponent is the number of terms, and this remainder multiplied by the first term. Hence, when we have given the sum of all the terms, the number of terms, and the ratio, to find the first term, we have this RUL E. Multiply the sum of all the terms by the ratio, less one; divide the product by the power of the ratio, whose index is the number of terms, after diminishing it by one. 240 HIGHER ARITHMETIC. E X A M P I, E S. 1. The sum of all the terms of a geometrical progression is 262143, the number of terms is 9, and the ratio is 4. What is the first term? In this example, the sum of all the terms, multiplied by the ratio, less one, is 262143 x 3= 786429; the power of the ratio, whose exponent is the number of terms, is 49 =262144; this, diminished by 1, becomes 262143;.'. 786429, divided by 262143, gives 3 for the first term. 2. The sum of all the terms of a geometrical progression is 591 7 41, the number of terms is 7, and the ratio is 4. What is the first term? Ans. 9. 3. If a debt of $4095 is discharged in 12 months by paying sums which are in geometrical progression, the ratio of which is 2, how much was the first payment? Ans. $1. Case VII. We have shown under Case II. that the sum of all the terms, multiplied by the ratio, less one, is equal to the first term subtracted from the last term into the ratio; therefore, the first term is equal to the product of the ratio into the last term, diminished by the product of the ratio, less one, into the sum of all the terms. Hence, when we have given the sum of all the terms, the last term, and the ratio, to find the first term, we have this RULE. Multiply the last term by the ratio, and from the product subtract the product of the sum of all the terms into the ratio, less one. GEOMETRICAL PROGRESSION. 241 EXAMPLES. I. The sum of all the terms of a geometrical pro. gression is 436905, the last term is 327680, and the ratio is 4. What is the first term? In this example, we find the last term, multiplied by the ratio, to be 1310720. The product of the sum of the terms into the ratio, less one, is 1310715;'. 1310720 - 1310715 = 5, for the first term. 2. The sum of all the terms of a geometrical progression is 6138, the last term is 3072, and the ratio is 2. What is the first term? Ans. 6. 3. The sum of all the terms of a geometrical progression is 1860040, the last term is 1240029, and the ratio is 3. What is the first term? Ans. 7. Case VIII. From the condition under Case II., we see that the ratio, multiplied into the sum of all the terms, diminished by the last term, is equal to the sum of all the terms, diminished by the first term. Hence, when we have given the first term, the last term, and the sum of all the terms, to find the ratio, we have this RULE. -Divide the sum of all the terms, diminished by the first term, by the sum of all the terms, diminished by the last term. EXAMPLES. 1. The first term of a geometrical progression is 5, the last term is 327680, and the sum of all the terms is 436905, What is the ratio? 21 242 HIGHER ARITIIMETIC. In this example, the sum of all the terms, diminished by the first term, is 436900, and the sum of all the terms, diminished by the last term, is 109225;.'. 436900, divided by 109225, gives 4 for the ratio. 2. The first term of a geometrical progression is 6, the last term is 3072, and the sum of all the terms is 6138. What is the ratio? Ans. 2. 3. The first term of a geometrical progression is 7, the last term is 1240029, and the sum of all the terms is 1860040. What is the ratio? Ans. 3. NOTE.-The demonstration of the rules for the four following cases have not been given; they may, however, be obtained by combining the conditions of some of the foregoing cases. Case IX. Given the first term, the ratio, and the sum of all the terms, to find the last term. RULE. To the first term add the product of the ratio, less one, into the sum of all the terms; divide this sum by the ratio. EXAMPLES. 1. The first term of a geometrical progression is 4, the ratio is 3, and the sum of all the terms is 118096. What is the last term? In this example, the product of the ratio, less one, into the sum of all the terms is 236192, which, added to the first term, gives 236196; this, divided by the ratio, gives 78732, for the last term. 2. A man bought a certain number of yards of cloth, GEOMETRICAL PROGRESSION. 243 giving 3 cents for the first yard, 6 cents for the second yard, 12 cents for the third yard, and so on, for the succeeding yards. If the whole number of yards cost $122-63, what did the last cost? Ans. $62'33. 3. A person bought a certain number of pears for ~4 5s. 3d. 3far.; he gave 1 farthing for the first, 2 farthings for the second, 4 for the third, and so on, doubling each time. What did he pay for the last? Ans. ~2 2 s. 8 d. Case X. Given the ratio, the number of terms, and the sum of all the terms, to find the last term. RULE. Raise the ratio to a power whose exponent is the number of terms, less one; multiply together this power, the sum of all the terms, and the ratio, less one; then divide this product by one less than the power of the ratio, whose eaxponent is the number of terms. EXAMPLES. 1. The ratio of the terms of a geometrical progression is 3, the number of terms is 10, and the sum of all the terms is 1 8S096. What is the last term? In this example, the ratio, raised to a power whose exponent is the number of terms, less one, is 39 =19683; this, multiplied by the sum of all the terms, and the ratio, less one, is 19683 x 1 18096 x 2 _ 4648967136; the power of the ratio, whose exponent is the number of terms, is 59049; this, diminished by 1, becomes 59048;.'. 4648967136, divided by 59048, gives 78732, for the last term. 244 HIGHER ARITHMETIC. 2 The ratio of the terms of a geometrical progression is 3, the number of terms is 10, and the sum of all the terms is 295240. What is the last term? Ans. 196830. 3. The ratio of the terms of a geometrical progression is 2, the number of terms is 11, and the sum of all the terms is 20470? What is the last term? Ans. 10240. Case XI. Given the first term, the number of terms, and the last term, to find the sum of all the terms. RULE. Extract the root denoted by the number of terms, less one, of the last and first terms; then raise these roots to a power, whose exponent is the number of terms; then divide the difference of these powers by the difference of the roots. EXAMPLES. 1. The first term of a geometrical progression is 1, the number of terms is 10, and the last term is 19683. What is the sum of all the terms? In this example, we must extract the 9th root of the last and first terms, which give 3 and 1 for the roots; these must each be raised to the 10th power, which give 59049 and 1, the difference of which is 59048; this, divided by 3-1= 2, gives 29524, for the sum of all the terms. 2. The first term of a geometrical progression is 1, the last term is 2048, and the number of terms is 12. What is the sum of all the terms? Ans. 4095. GEOMETRICAL, PROGRESSION. 245 3. The first term of a geometrical progression is 1, the last term is 10077696, and the number of terms is 10. What is the sum of all the terms? Ans. 12093235. (ase XII. Given the ratio, the number of terms, and the last term, to find the sum of all the terms. RULE. Raise the ratio to a power whose exponent is the number of terms; from this power subtract one, and multiply the remainder by the last term; divide this product by the product of the ratio, less one, into the power of the ratio, whose exponent is the number -of terms, less one. EX AMPLES. 1. The ratio of the terms of a geometrical progression is 2, the number of terms is 12, and the last term is 2048. What is the sum of all the terms? In this example, the ratio, raised to a power whose exponent is the number of terms, is 212 =4096; this, diminished by 1, becomes 4095, which, multiplied by 2048, becomes 8386560; again, the power of the ratio, whose exponent is one less than the number of terms, is 2048, which, multiplied by the ratio, less one, is not changed;.. 8386560, divided by 2048, gives 4095, for the sum of all the terms. 2. The ratio of the terms of a geometrical progression is 3, the number of terms is 8, and the last term is 106 23. What is the sum of all the terms? Ans. 3074 41. 21' 246 HGIGHER ALt1TIIMETIC.'3. The ratio of the terms of a geomnetrical progression is 4-, the number of terms is 7, and tie last term is 2582 0 73. What is the sum of all the terms? Ans. 591 74. NOTE.-The eight remaining cases in geometrical progression cannot be solved by the ordinary processes of arithmetic, but require for their solution a knowledge of logarithms, and algebraic equations above the second degree. 83. WHEN the ratio of a geometrical progression is less than a unit, the first term will be the largest, and the last term the least; the progression, will, in this case, be descending. But, if we consider the series of terms in a reverse order, that is, calling the last term the first, and the first the last, the progression may then be considered as ascending. If a decreasing geometrical progression be continued to an infinite number of terms, we may neglect the last term as of no appreciable value; we can find the sum of such a progression by Case II., when it is modified, as follows: Given the first term of a descending geometrical progression, and the ratio, to find the sum of all the terms, when continued to infinity. RULE. Divide the first term by a unit diminished by the ratio. EXAMPLES. 1. What is the sum of all the terms of the infinite series 1, d, 4, 8, &c.? GEOMETRICAL PROGRESSION. 247 In this example, a unit, diminished by the ratio, is 1 —J-'; and the first term, 1, divided by 2, gives 2, for the sum of all the terms. 2. What is the sum of the infinite series 1, 3, a, 2, &c.? Ans. 1~2. 3 What is the sum of the infinite series,1 3v,200, o ~Ao,- &c.? Ans.. 4. What is the sum of the infinite series 2_, 4 5. What is the sum of the infinite series 1 -,1 0 o,3;09 o0ooo, &c.?? Ans. -1,. 6. What is the sum of the infinite series T1, TI6W o6 216 &c? Ans. 4. 248 HI GHIER A R ITIMETIC CHAPTER XII. MISCELLANEOUS QUESTIONS SOLVED BY ANALYSIS, 84. A MAN and his wife usually drank out a cask of beer in 12 days; but when the man was from home, it lasted the woman 30 days. How many days would the man alone be in drinking it'? Solution. Since it requires 12 days for the man and his wife to drink out the cask, they must, in each day, drink r2 of it. Again, since the woman is 30 days in drinking it, she must, in each, day, drink g-j of it. Hence, the fractional part which the man drank in 1 day, must be'-2 — =-2 as;:. in 20 days he could drink the whole. 2. A person bought several gallons of wine for $94, and after using 7 gallons himself, sold I of the remainder for $20. How many gallons had he at first? Solution. Since he sold I of the remainder, after using 7 galIons, for $20, he could have sold the whole of the remainder for $80; therefore, the value of the 7 gallons which he used was 94-80= —$14; and 1 gallon must have cost -4-=-$2. The wine being worth $2 per gallon, he must have purchased -2-=47 gallons. QUESTIONS SOLVED BY ANALYSIS. 249 3. A person in play lost 4 of his money, and then won 3 shillings, after which he lost 3 of what he then had; and this done, he found that he had but 12 shillings remaining. How much had he at first? Solution.'This, like all other questions of a similar nature, is most readily solved by an inverse operation. Thus, since the 12 shillings which finally remained, was only 2of what he had before losing 4, we may find what he had before losing the 4, by dividing 12 shillings by 2, which thus becomes 3 of 12 shillings —18 shillings; now, deducting the 3 shillings which he won, we get 15 shillings, which is 3 of what he had at first; therefore, 4 of 15 shillings-20 shillings, is what he had at first. 4. A fish was caught whose tail weighed 9 pounds; his head weighed as much as his tail and half his body, and his body weighed as much as his head and tail together. What was the weight of the fish? Solution. Since the head of the fish is equal to I of the body, together with the tail, which equals 9 lbs., it follows that the head and tail together must equal 4 of the body+ 18 lbs. But, by the question, the head and tail together is equal to the whole body;.'. we have this relation: I of the body+ 18 pounds, must equal the whole body; consequently, 4 of the body must equal 18 pounds, and the whole body is 36 pounds. And, since the body weighed as much as the head and tail together, it follows that the weight of the whole fish was twice that of the body, or 8 times that of the tail, which is 72 pounds. 250 HI It EIi ARI1'nET 1IC. 5. A person engaged a workman for 48 days. For each day that he labored he received 24 cents, and for each day that he was idle, he paid 12 cents for his board. At the end of the 48 days, the account was settled, when the, laborer received $5'04. Required the number of working days, and the number of days he was idle. Solutiomn, Had he worked all the time, he would have received 24x48=$11'52; but he received only $5'04. Therefore, by being idle, lie lost $11'52 —$5'04 -$6'48. Now, for each idle day, he loses the 24 cents which he might have earned, as well as the 12 cents which he gives for his board; so that every idle day is to him a loss of 24+12-36 cents. But we have just shown that his total loss was $6'48;.. the number of idle days was i-A-:8- 18, and he worked 48- 18 - 30 days. 6. A gentleman bought two pieces of silk, which, together, measured 36 yards. Each of them cost as many shillings per yard as there were yards in the piece, and their whole prices were as 4 to 1. What were the lengths of the pieces? Solution. Since each piece cost per yard as many shillings as there were yards in its length, it follows that their values, expressed in shillings, must be as the squares of their lengths. By the question, their prices were as 4 to 1; therefore, the squares of their lengths must be to each other as 4 to 1; consequently, their lengths must be to each other as 2 to 1. The question is now reduced to the following: divide QUESTIONS SOLVED BY ANALYSIS. 251 36 into 2 parts, which shall be to each other as 2 to 1. These parts are 23 of 36=24, and I of 36=12. 7. In a mixture of wine and cider, 2 of the whole, +25 gallons, is wine, and ~ of the whole,-5 gallons, is cider. How many gallons were there of each? Solution. By the question, the wine = - of the whole+25 galIons, and the cider=- of the whole-5 gallons. Hence, taking the sum of these expressions, we get the whole -=(.+~) or A of the whole + 20 gallons;.-. of the whole equals 20 gallons; consequently, the whole is 120 gallons. Now I of the whole is 60 gallons, to which, add 25 gallons, we get, for the wine, 85 gallons. Again, I of the whole is 40 gallons, from which, subtracting 5 gallons, we get, for the cider, 35 gallons. 8. A market-woman bought a certain number of eggs, at 2 for a penny, and as many more at 3 for a penny; and, having sold them again, all together, at the rate of 5 for 2 pence, found that she had lost 4 pence. How many eggs had she? Solutioan. Since, by the question, half of the eggs cost I of a penny apiece, and the other half cost I of a penny apiece, it follows that the average price which she gave for the eggs was ( + )- -= -5- of a penny apiece. Since she sold them all together at the rate of 5 for 2 pence, that is, 2 of a penny apiece, she must have lost on each egg 15 2 - = of a penny. Therefore, to lose 1 penny, she must have disposed of 60 eggs, and to lose 4 pence, she must have had 240 eggs. 252 HIGHER AR I'IIMETIC. 9. A and B can, together, do a piece of work in 8 days; A and C can, together, do it in 9 days; B and C can, together, do it in 10 days. How many days would it require for each to perform the work alone? solutiot. Since A and B can do the work in S days, they can, in 1 day, do 8 part of it; for a similar reason, A and C can do - part of it in I day; B and C can do,- part of it in 1 day. Adding these fractional parts together, and observing that each individual has been included twice, we shall get, by dividing the sum by 2, the following fraction, (I + I + -). 24, — which is the fractional part of the work which they all together would perform in 1 day. We have already seen that the part which B and C can perform in 1 day is -;.J. - o —- -o, is the fractional part which A could perform in 1 day. Hence, the time in which A could alone perform the work is -7290 -=1439 days. Again, the fractional part which A and C together could perform, is'1.- 72-I — 4- is the fractional c pr 7 2 0 9 -- 7 2O0, part which B conld perform in 1 day; hence, the time in which B could alone perform the work, is _74-2_172 17days, The fractional part which A and B together could perform, is,;.-. 2- I-=, o, is the fractional part which C could perform in I day; hence, the time in which he' could alone perform the work is -L1- -=23-t- days. 10. A and B have each the same income. A contracts an annual debt, amounting to 7 of it; B lives on 4 of it and, at the end of ten years, B lends to A enough to pay off his debts, and has $160 left. What is the incomne? tiESTq'IONS SOLVED) LBY ANAL IS. 253 Since B lives on - of his income, he must save 5 of it. A's debt for 1 year being 4 of the income, B will have left, after paying A's debt, _5- = -52 of his income. And, since this would in 10 years amount to $160, -P of his income must equal $16. Hence, the income was 23_5 of $16= —$280. 11. A merchant supported himself 3 years for $50 a year; at the end of each year, he added to that part of his stock which was not thus expended, a sum equal to 4 of this part. At the end of the third year, his original stock was doubled. What was the stock? Solutions. After supporting himself the first year, lie will have his original stock, -$50; this, increased by its third part, will become 4 of his original stock, -- of $50; living upon another $50, he will have left 4 of his original stock, -4 of $50-$50. This must again be increased by its third part, giving X6- of his original stock, - 6- of $50- 4 of $50. Again, living upon $50, he will have left -16 of his original stock, - --- of $50- 4 of $50- $50; increasing this once more by its third part, we get 64 of his original stock, -64 of $50 —'6 of $50-4 of $50. This, by the question, is equal to twice his original stock, or to 247 of his original stock. Hence, - - 4= of his original stock must equal _4 of $50+ +-L of $50+ 3 of $50=-l-4-8 of $50=-424dollars;.. his stock was 7 4_ 1 o_- $740. 12. Fourteen oxen have, in 3 weeks, eaten all the 22 254 1 G i ER AR ITH E TI C. grass which grew on 2 acres of land, in such a manner, that they not only ate all the grass which at first was there, but also that which grew during the time they were grazing. In like manner have 16 oxen, in 4 weeks, eaten all the grass upon 3 acres of land. How many oxen can, ill this way, graze for 5 weeks upon 6 acres of land? S'olutio,0, If the grass of 2 acres, Acres. Oxen. Weeks (1.) 2 14 31 with its growth for 3 wks., (2.) 1 7 3 keep 14 oxen for 3 weeks, (3.) 3 79 B19st condition. then will the grass of one (4.) 3 63 1 J acre, with its growth for (5.) 3 16 4 3 weeks, keep 7 oxen for (6.) 3 64 1 By2condition. (6.) 6 3 weeks; and the grass of 3 acres, with its growth for 3 weeks, will keep 7 oxen 9 weeks, or 63 oxen for 1 week, which result corresponds with (4) in the adjoining table. Again, by the second condition of the question, if the grass of 3 acres, with its growth for 4 weeks, keep 16 oxen for 4 weeks, then will the grass of 3 acres, with its growth for 4 weeks, keep 64 oxen for 1 week, which corresponds with (6). By carefully comparing (4) and (6), we see that tile growth of 3 acres for 1 week is sufficient tc keep 1 ox 1 week; consequently, the growth of 6 acres for 1 week will keep 2 oxen for 1 week; or, which is the same thing, the growth of 6 acres for 5 weeks will keep 2 oxen for 5 weeks. By (4) we have seen that the grass of 3 acres, with its growth for 3 weeks, will keep 63 oxen for 1 week; but the growth of 3 acres for 3 weeks, will keep 3 oxen QUESTIONS SOLVED BY ANALYSIS. 255 1 week; consequently, the grass alone of 3 acres will keep 60 oxen 1 week, and the grass of 6 acres will keep 120 oxen for 1 week, or it will keep 24 oxen for 5 weeks. Hence, finally, the grass of 6 acres, with its growth for 5 weeks, will keep, during 5 weeks, 2+24=26 oxen. 13. A person expends just ~100 for live stock, consisting of geese, sheep, and cows; for each goose he paid 1 s., for each sheep ~1, and for each cow ~5. How many did he purchase of each kind, so as to have just 100 in all? Solution. Since the average price of the animals was ~1 =20s., the price of a goose was 19 s. below the average, and the price of a cow was 80 s. above the average. Hence, were he to purchase the geese and cows only in the ratio of 80 geese to 19 cows, their whole cost would be just as many pounds sterling as there were animals; thus, by purchasing 80 geese and 19 cows, he would have 99 animals, together worth ~99. Now, by adding 1 sheep, worth ~1, he will have 100 animals, together worth ~100. So that he bought 80 geese, 1 sheep, and 19 cows. 14. A and B leave Utica for Albany at the same time that C leaves Albany for Utica. If A goes 8 miles each hour, B 12 miles, and C 9, when will C be equally distant between A and B, if the distance between Utica and Albany is 95 miles? Solution. The average velocity of A and B is 10 miles an hour; hence, if D, a fourth person, leave Utica at the same time, going 10 miles each hour, he will always be equally 256 HIGHER ARITHMETIC. distant between A and B, and the time sought will be when he is met by C. Now D and C together travel 19 miles each hour; consequently, they will meet in 95. 19=5 hours, at which time C will be equally distant from A and B. 15 A, B, C, D, and E, play together on this condition: that he who loses shall give to all the rest as much as they already have. First, A loses, then B, then. C, then D, and at last also E. All lose in turn, and yet, at the end of the fifth game, they have all the same sum, viz., each $32. How much had each before they began to play? Solution. The solution of this question is the most readily effected by a reverse process; that is, by beginning with the last game, and playing them all in a reverse order, as follows: first, take from A, B, C, and D, half they have, and add it to E's money; second, take from A, B, C, and E, half what they now have, and add it to D's; third, take from A, B, D, and E, half what they now have, and add it to C's; fourth, take half of A's, C's, D's, and E's, and add it to B's; lastly, take half of B's, C's, D's, and E's, and add it to A's. These successive operations may be exhibited as follows: A. B. C. D. E. $32 $32 $32 $32 $32, end of 5th game, 16 16 16 16 96, end of 4th game, 8 8 8 88 48, end of 3d game, 4 4 84 44 24, end of 2d game, 2 82 4A. 22 12, end of 1st game, Ans. 81 41 21 11 6, before playing. qUESTIONS SOLVED BY ANAI,YSIS. 257 16. A father left to his thriee sons, whose ages are 8, 10, and 13 years, $10000, to b)e so divided that the respective parts, being placed out at 5 per cent., compound interest, should amount to equal sums when they became 21 years of age. What are the parts? Solution. By tile question, their respective shares would be at interest, 13, 11, and 8 years. We find, by table under Art. 69, the present worth of $1 for 13, 11, and 8 years respectively, at 5 per cent., compound interest, to he $0'530321, $0'584679, and $0'676839. Now it is obvious that the parts must be to each other in the same ratio as the numbers 530321, 584679, and 676839; the sum of these numbers is 1791839. Hence, the parts are as follows: The 1st one's part is 3-5-3-12 of $10000=- $2959'646. The 2d one's part is,5 8-4 679 of $10000=-$3263'011. The 3d one's part is -67 6883s9 of $10000=$3777'343. 17. Find what each of the four persons, A, B, C, and D, are worth, by knowing, 1st. That A's money, together with I of B's, C's, and D's, is equal to $137. 2d. That B's money, together with I of A's, C's, and D's, is equal to $137. 3d. That C's money, together with I of A's, B's, and D's, is equal to $137. 4th. That D's money, together with i of A's, B's, and C's, is equal to $137. Soluhion. It is evident that - of B's, C's, and D's money is the same as ~ of the sum of all, MINUS 3 of A's; therefore, 22* 258 HIiIGHER ARITHMETIt'. A's. together with I of B's, C's, and D's, is equal to A's + I of the sum of all — I of A's, which, by the first condition, equals $137. Consequently, 2 of A's $137 — of the sum of all... A's= 3 of $137- of the sum of all. In a similar way, we get B's= 3 of $137-f - of the sum of call. C's=-5- of $137 — I of the sum of all. D's=-6 of $137 — of the sum of all. Taking the sum of these values of A, B, C, and D, we get the sum of all= —( 3 + 4 q -5) of $137-( 2 -+ ~ ~ + -) of the sum of all... (1 + + ) of the sum of all=(f3-+4+- 5+6) of $137. And the sum of 3 4 1, 5 1 6 all= 2 —3~4~5 of $137= 31- of $137=$317.' 1 A-f 1 31 ~ f 2 3 4 5 This value for the sum of all, being substituted in the above values of A, B, C, and D, we obtain the following results: A's=3 of $137- of $317 —$ 47. B's-=4 of $137- 1 of $317 —$ 77. C's= -of $137 — of $317= —$ 92. D's= - of $137- of $317=$101. NOTE.-It is obvious that this method of solving the above question will apply in the case of any number of unknown quantities which are similarly related to each other. 18. A and B, settling accounts, found that if ~6 were added to 2- of A's money, and the same sum taken from -3 of B's, the sum would be 7 of the remainder, and that the sum and the remainder, added together, made ~72. What was each man's money? Solution,. Since ~6 was added to - of A's money, and subtracted QUESTIONS SOLVED BY ANALYSIS. 259 from 2 of B's money, the sum of A's and B's money, after this addition and subtraction, is the same as it would have been had no such addition and subtraction been made. Therefore, z of A's and B's money is, by the question, ~72. Again, by the question, 23 of A's, increased by ~6, is equal to 7 of 2 of B's, diminished by ~6;.. 2 of A's, increased by ~6, is to 2 of B's, diminished by ~6, as 7 to 8. But we have already seen that 2- of A's, increased by ~6, added to 2 of B's, diminished by ~6, is ~72. Hence, if we divide ~72 into two parts which are to each other as 7 to 8, these parts will be 2- of A's, increased by ~6, and 2 of B's, diminished by ~6. The parts of ~72 which are to each other as 7 to 8, are x5 of ~72-=333, and -8- of ~72-~-382. Therefore, -23 of A's, increased by ~6, is equal to ~33-; consequently, 2 of A's is ~333-~6=~273; and the whole of A's money is 3 of ~27 —=~41~. And 2 of B's, diminished by ~6, is equal to ~382. Therefore, - of B's is ~382 +~6=~442-; and the whole of B's money is 2 of ~442 =~663. 19. A purse of $2850 is to be divided among three persons, A, B, and C. A's share is to be to B's as 6 to 11, and C is to have $300 more than A and B together. What is each one's share? Solution. Since C is to have $300 more than A and B together, it follows that A and B must have half of what is left, after subtracting $300; hence, A and B together have $1275; this, divided into two parts, which are to each other as 6 to 11, gives A's= —16 of $1275-$450; B's= ~. of $1275 =$825; C's is evidently $1575. 260 HIGHER ARITHMETIC. 20. Two persons, A and B, purchase in company 100 acres of land for $1000, of which the southern portion is of rather the best quality. In the division of it, A says to B, let me have the southern portion, and I will pay for my part $1 -- per acre more than you pay for yours. How much land must each have, and at what price per acre? NOTE.-This, and like questions, can be solved by the following RULE. Divide half the whole cost by the whole number of acres, and to the square of the quotient add the square of half the difference of the prices per acre; then extract the square root of the sum, and to this root add the quotient of half the whole cost, divided by the whole number of acres. This last sum, increased by half the difference of the prices per acre, will give the price per acre of the best land; and, diminished by the same, will give the price per acre of the poorest. Applying the above rule, we find the quotient of half the whole cost, divided by the whole number of acres, to be 5, which, squared, gives 25; this, increased by the square of half the difference of the price per acre, becomes 25+ 6- 31276_1 whose square root is 138a; this root, added to 5, gives -3 —l= 10k-. Therefore, the price per acre of A's land, is 10l+~ 9_=$105. The price of B's land is 10. _J-= $9. $500 $105 =47-1 acres, for A's portion. $500 $ 9- = 52- 2 acres, for B's portion. 21. A boy divided his apples among his four companions in the following manner: To the first he gave half an apple more than half his whole number; to the second he likewise gave half an apple more than half the number which he then had; in the same manner he QUESTIONS SOLVED BY ANALYSIS. 261 divided with the third and fourth companion, giving to each half an apple more than half the number which was left after giving to the preceding one. After having divided with the fourth companion, he had but one apple left. How many had he at first? Solutiown. This question is most easily solved by beginning with the last companion, and reversing each successive operation. Since he had but one apple left after making the last division, he must have had 3 after the third division, because 3, diminished by I more than its half, leaves 1; for a similar reason, he must have had 7 after the second division. In this way we may retrace the process, by multiplying the number by 2, and adding 1 to the product for each successive step. In this way we find that, after the first division, he must have had 7 x 2 +1= 15. And, before he divided with the first, he must have had 15x2+1=31 apples. NOTE. —From the above method of solving this question, we see that it would not be difficult to extend the solution to the case of any number of divisions. Indeed, we see that the number required is always one less than 2, raised to a power whose index is one more than the number of divisions. Thus, had there been 10 companions to divide with, after the above method, he must have had 21- -1=2047 apples at first. This method of division is effected without dividing an individual apple. 22. A hare is 50 leaps before a greyhound, and takes 4 leaps to the greyhound's 3, but 2 leaps of the hound are equal to 3 of the hare's. How many leaps must the greyhound take before he catches the hare? 262 HIGHER ARITHMETIC. Solution. Since 2 leaps of the greyhound equals 3 leaps of the hare, it follows that 6 leaps of the greyhound equals 9 leaps of the hare. But, while the greyhound takes 6 leaps, the hare takes 8 leaps; therefore, while the hare takes 8 leaps, the greyhound gains on her 1 leap. Hence, to gain 50 leaps, she must take 50 x 8= —400 leaps; but, while the hare takes 400 leaps, tile greyhound would take 300 leaps, since the number of leaps taken by them were as 4 to 3. 23. From a cask of wine a tenth part is drawn out, and then the cask is filled with water; after which, a tenth part of the mixture is drawn out; again the cask is filled, and again a tenth part of the mixture is drawn out. Now suppose the fluids mix uniformly at each time the cask is replenished, what fractional part of wine will remain after the process of drawing out and replenishing has been repeated ten times? Solution. Since -lo of the wine is drawn out at the first drawing, there must remain ~-. And after the cask is filled with water, -, of the whole being drawn out, there will remain -~, of the mixture; but ~-9o of the mixture, we have already seen, is wine; therefore, after the second draw92 Ing, there will remain,L- of 7-0% of wine, or -. By similar reasoning, we see that after the third drawing 93 there will remain 9r- of -r9 of oo of wine, or 103 QUESTIONS SOLVED BY ANALYSIS. 263 From this, we see that the part of wine remaining is expressed by the ratio -?-, raised to a power whose exponent is the number of times the cask has been drawn from. IHence, in the present question, the fractional 910 part of wine is 90 3-04 -06-8o40_ =7 0-3486784401, p a r If 1010o0 0000 which is nearly 35 per cent. 24. There are two vessels of equal capacity, the one 2 full of wine, the other 2- full of water; now suppose the vessel containing the wine to be first filled from the water vessel, and then the water vessel to be filled from the wine vessel; and so on, alternately filling the one from the other. Now, on the supposition that the fluids mix uniformly at each time, how much wine, and how much water will the wine vessel contain after it has been filled 5 times? And how much water, and how much wine will the water vessel contain after it has been filled 5 times? Solut;ion. In the first place, the wine vessel is filled from the water, so that it will consist of 9- of wine, and ~ of water, and there will remain in the water vessel ~ of water. Now the water vessel is filled from the wine vessel by drawing from it 2; that is, 2 of 2- of wine=4 of wine, and 2- of ~ of water= - of water, so that the water vessel consists of.- of water and A of wine; and there remains in the wine vessel 2- of wine, and - of water. We next fill the wine vessel, and find that it will contain -47 of wine, and 2 of water, and that there will remain in the water vessel J- of water, and -A of wine. 264 1; i I:1 A.R ITlMiTC. These successive operations will be more clearly comprehended by the aid of the following TABLE. WATER. WINE. WATER. WINE. 2 i 2 3 3 2 Wine vessel 3Vate ve sel filled 1st time. Water vessel filled 1st time. 5 4 2 1 -25 4A 1 3 I4 Wine vessel _ _ _2 __2_7_ 2 2__ _ _ _ _ filled 2d time. Water vessel!! filled 2d time. 4 1 440 1 4 88 1 4 1 4 0 121 2 2 Wine vessel 2_4_3 2_4_ 3 2_4_ 2__ 4_3 filled 3d time. Water vessel filled 3d time. 3 6 5 2 1 1 2 2 I 36 5 3 6 4 1 0193 1O94 T 5T 218- 7 2 187 2A 9 Wine vessel filled 4th time. Water vessel filled 4th time. 3 2 8 1 3 2 8 0 1 0 9 3 1 0 9 4 65671 656 1 656 1 1 656 1.iled thtimefilled 5th time. fiater vessel filled 5th time. 29525 29524 984 1 9 842 5 9 0 4 9 5 9 0 4 9.5 9049 I 5 9T04 9 The law of continuation is so simple, that this table might with ease be carried to any extent desired. The QUESTIONS SOLVED BY ANALYSIS. 265 denominators of the partial fractions expressing the wine and water, are the successive powers of 3, and the numerators are found by dividing the denominators into two parts, differing from each other by a unit. Thus, after the 5th pouring, the wine vessel has been 1(35 1) filled 3 times, and the quantity of wine is 2(3 1) 35 -2- and the water is 2 1)=21 After the 6th, = a4 3 A3 pouring, the water vessel has been filled 3 times, and the water is 2(3+-1)_65 and the wine is (3-1) -362 36 _364 We may remark that an odd number of pourings leaves the wine vessel full, and, consequently, the water vessel 3 full; and an even number of pourings leaves the water vessel full, and, consequently, the wine vessel - full. We may also remark that the fluids will never become equally mixed, since the denominators which express the fractional parts of wine and water, at any particular time, are always the same, while their numerators differ a unit; so that the wine vessel will always contain an excess of wine, and the water vessel an excess of water. Notwithstanding they are never equal, still they approach more nearly to an equality the greater the number of times the operation is performed. 25. A person possesses a wagon with a mechanical contrivance, by which the difference of the number of revolutions of the wheels on a journey may be determined. It is known that each of the fore wheels are 54 feet, and that each of the hind wheels are 7- feet in 23 266 HIGHER ARITHMETIC. circumference. Now, when on a journey, the fore wheel has made 2000 revolutions more than the hind wheel; how great was the distance traveled? Solution. Circumference of fore wheel - 5 ft. = - = —- feet. hind wheel = 7- ft. - 5- feet. Now find the least number that will divide by 42 and 57, (by Rules under Art. 10 or 11.) This number is 798. Hence, the least number which will divide by -48and -7 is 798 Now, in going * feet, the fore wheel turns over 7 98 42 =19 times; and the hind wheel turns over 789 58-= 14 times; therefore, in making a journey of 7 8' feet, the fore wheel turns over 19 —14=5 times more than the hind wheel. Hence, in order that the fore wheel may turn over 2000 times more than the hind wheel, he must perform a journey of 7 9 x 2 o-3~ = 39900 feet, the answer. 26. Suppose $1 had been put out at compound interest, at 7 per cent., the 5th day of October, 1585, how much would it have amounted to on the 5th day of October, 1841? Solution. By referring to what has been said under the subject of compound interest, we see that we must find the 256th power of 1'07. After reaching the 4th power, the subsequent multiplications have been performed by the abridged method, Art. 39. QUESTIONS SOLVED BY ANALYSIS. 267 2 952163749 1'07= 1 year. 2'952163749 1'07 5904327498 749 2656947374 107 147608187 1'1449 = 2 yrs. 5904327 1P1449 295216 103041 8856 45796 2066 45796 118 11449 26 11449 8' 715270798 32 yrs. 1'310796010 =4 yrs. 8715270798 1'310796010 69722166384 1310796010 6100689559 393238803 87152708 13107960 43576354 917557 1743054 117972 610069 7865 6101 13 784 l'718186180 = 8 yrs. 70 1'7181 6180 75'955945083 = 64 yrs. 1718186180 75'955945083 1202730326 531691615581 17181862 37977972541 13745489 6836035057 171819 379779725 137455 37977973 10309 6836035 172 303824 137 37978 2952163749 = 16 yrs. 608 23 5769'305593425 = 128yrs. 268 IIIGHE R ARI' II METIC. 5769'30559345 5769'30559345 2884652796725 403851391542 34615833561 5192375034 173079168 2884652 51924 1731 231 29 Ans. $33284887'03062=- 256 years. 27. Suppose from an acorn there shoots up a single stalk, at the end of the year; that, at the end of each year thereafter, this stalk puts forth as many new branches as it is years old; also suppose all the branches to follow the same law, that is, to produce as many new branches as they are years old. How many branches will this oak tree consist of at the end of 20 years. Solution.* From the conditions of the question, we know that at the end of the first year there will be simply one stalk, or branch, which we will denote by 1 o; at the end of the second year, this branch will become 1 year old, and produce a new branch, so that we shall have 1 + 1 o; at the end of the third year, the branches, 1, + 1, will become 12 + 1, the first of which being 2 years of age, wig produce 2 new branches; the other will produce * For an algebraic solution of this, see Recurring Series in my ALGEBRA. QUEST'IONS SOLVED BY ANALYSIS. 269 1 new one; we shall, therefore, have 12 + 11 + 3,,. Proceeding in this way, we obtain the following results: End of 1st year,.. 1 "( 2d " ll+lo 1 2 3d " 12 + 1 1+30 5 4th " 13+12+31+8o =-13 5th " 14+13+3 2+81 +210 =34 "' 6th " 15214+33+82+211+-550=-89 In this scheme, the small figures at the bottom of the larger ones, denote the age, in years, of the branches to which they are attached. Thus, at the end of the fifth year, there will be 1 branch 4 years old, 1 branch 3 years old, 3 branches 2 years old, 8 branches 1 year old, and 21 new branches of no age. The law of the above series is obvious. It is such, that twice any term, increased by the sum of all the preceding terms, gives the next succeeding term. Or, three times any term, decreased by the next preceding term, will give the next succeeding term. These terms may be found most easily by continual addition, as given by the following table, where each succeeding term is found by adding the two preceding ones. 0 New branches 1st year. Total branches 1st year,. 1 1 New branches 2d year. Total branches 2d year, 2 3 New branches 3d year. Total branches 3d year,. 5 8 New branches 4th year. Total branches 4th year, 13 21 New branches 5th year. Total branches 5th year,..34 55 New branches 6th year. Fotal branches 6th year,. 89 23* 270 GIIItIHER ARITIIiMET''IC. 144 New branches 7th year. Total branches 7th year,.. 233 377 New branches 8th year. Total branches 8th year... 610 987 New branches 9th year. Total branches 9th year,.. 1597 2584 New branches. 10th year. Total branches 10th year,.. 4181 6765 New branches 11th year. Total branches 11th year,. 10946 17711 New branches 12th year. Total branches 12th year,. 28657 46368 Nemw branches 13th year. Total branches 13th year, 75025 121393 New branches 14th year. Total branches 14th year,. 196418 317811 New branches 15th year. Total branches 15th year,. 514'229 832040 New branches 16th year. Total branches 16th year,. 1346269 2178309 New branches 17th year. Total branches 17th year,.3524578 5702887 New branches 18th year. Total branches 18th year,.9227465 14930352 New branches 19th year. Total branches 19th year, 24157817 39088169 New branches 20th year. Total branches 20th year, 63245986 28. Find ten numbers, such that the first, with 5 of all the others, shall make 845693; the second, with,3 of all the others, shall make 845693; the third, with 7 of all the others, shall make 845693; the fourth with o, the fifth with a, the sixth with -,, the seventh with -T, the eighth with -, the ninth with -, and the tenth with TT of all the others, shall make, respectively, the same number, 845693. SolIution. This may be solved by a method analogous to the one employed for the solution of question 17, pages 257-8. QUESTIONS SOLVED BY ANALYSIS. 271 Thus, 5 of the 2d, 3d, 4th, 5thll, 6th, 7th, 8th, 9th, and 10th, is the same as 5 of the sum of all, MINUS 5 of the 1st. Therefore, the 1st, together with 5 of the others, is equal to the 1st + - of the sum of all - of the 1st, which, by the first condition of the question, equals 845693. Consequently, 4 of the st = 845693-5 of the sum of all. Therefore, the lst — of 845693 — of the sum of all. In a similar way, we get, 2d- of 845693 — of the sum of all. 3d - 7 of 845693 — I of the sum of all. 4th=- of 845693 — of the sum of all. 5th-= 9 of 845693- 8 of the sum of all. 6th=-j — of 845693 — I of the sum of all 7th = 1 of 845693 —no of the sum of all. 8th= 2 of 845693- 1- of the sum of all. 9th- 1 of 845693- of the sum of all. 10th= — i of 845693 —,-A of the sum of all. If we take the sum of the above values, and observe that 5 + 6 + 7 + 8 _-LI- I 1+2 13 1 4 40 8 8 9 3 3 4 5 7 8'9 10 1 11 1 2 1 1- 3+O 36-0 4+ I+ I+ I+I I + _ _ + _+ _ + - + 3 3 I~ +TT T IT "6 ~'"', we shall obtain, Suim of all=- 4~8o93 of 845693 o- 03 6 of the sum of all. Trherefore, (+ 4 8 36 3 -) of the sum of all = -4 —-8-333 of 845693, or 84 5 69- 3 of the sum of all= 4o-08933 of 845693. Consequently, the sum of all —4088933. HavingT found the s.utm rf all, we can readily find the HII I 1GIE A R I liM E TiC. respective numbers by substituting it in tile hfregoing expressions, thus: lst= - of 845693 — 4 of 408933= 348$3. 2d -= of 845693 — I of 4088933=197045. 3d -= of 845693 — of 4088933=305153, 4th= 7 of 845693 — 4 of 4088933=382373, 5th= 9 of 845693- 4 of 4088933=440288, 6th=-P — of 845693- 4 of 4088933=485333. 7th —= of 845693 —l of 4088933=521369. 8th = 2 of 845693T-,T of 4088933 = 550853, 9th- -3 of 845693- - of 4088933 = 575423. 10th — = 4 of 845693 - of 4088933 = 596213. If, instead of the constant number 845693, it were required to have the respective numbers, when increased by the above fractional parts of the others, equal to any other constant number, we must muItiply these answers by the ratio of this new number to 845693. [For a complete algebraic solution of this, and all other analogous questions, see my ALGEBRA.] MiSCUELLANEOUS tQUESTIONS. 273 CHAPTER XIII. MISCELLANEOUS QUESTIONS. 85. WHAT are the prime factors of 2006? Ans. 2006=- 2 X 17 x 59. 2. vL.A- are the prime factors of 3742? Ans. 3742= 2 x 1871. 3. What is the greatest common measure of 720, 360, and 180? Ans. 22 X32 X5-180. 4. What is the greatest common measure of 420, 147, and 210? Ans. 3x 7=21. 5. What is the least common multiple of 4, 16, 24, and 40? Ans. 24 x 3 x 5 =240. 6. What is the least common multiple of 8, 36, and 100? Ans. 23 x 32 x 52 = 1800. 7. What are all the divisors of 376? Ans. 1, 2, 4, S, 47, 94, 188, 376. S. What are all the divisors of 3456? Ans. i 1, 2, 4, 8, 16, 32, 733, 1466, 2932, 5864, 11728, 23456. 9. What is the number of the divisors of 7866? Ans. 2x3x2x2=24. 10. What is the number of the divisors of 1000? Ans. 4x4==16. 11. Reduce 3 " 8 to its lowest terms. Ans.,}. 12. Reduce 234 4,8 to its lowest terms. Ans. 15 32 29321* 274 H I G H E AR A R XIt R E ti 1 C 13. Reduce the improper fraction 2 3-d to a mixed fraction. Ans. 1932. 14.- Reduce 4-67 to a mixed fraction. Ans. 6 3. 15. Reduce 67r-5- to an improper fraction. As. 4 742 16. Reduce 373 to an improper fraction. Ans. 13 2 17. What is the product of — 365y- into 3? Ans. 354 o 18. What is the product of 2, into loQ_? Ans. o. 19. Reduce the compound fraction 2 of 2 of 7 of 14 to a simple fraction. Ans. -'10 20. Reduce -:, -343, and 6, to fractions having a cornmon denominator. ns. Ans. 2' 21. What is the sum of 4 6 and 1? Ars 1 562 22. What is the quotient of -70 divided by -? Ans. 323. Reduce the complex fraction 4_ to its simplest form. Ans, 2 9 24. What is the value of,-3a of a mile? Ans. 1 fur. 16 rods, 7 - feet. 25. Can the vulgtlr fraction 3T be accurately expressed in decimals? Ans. It cannot. 26. How many places of decimals will be required to express -1? Ans. 4 places. 27. Find the compound repetend equivalent to - Ans. 0'038461.5 28. Find the perfect repetend arising from 2-. Ans. 0'02127659574468085106382 97872340425531914893617. 29. Convert 0'3756 into a vulgar fraction. Ans. 939 YTTWI MISCELLANEOUS QUESTIONS. 275 30. Convert 37- into a continued fraction. ns.1 1.+1 1+i -1 2-1 1+1 1+1 7+1 2. 31. Find some of the approximative values of the continued fraction 3+1 3+1 3+, &c. Ans. o, -3 1 ~ -33- C 32. What must be the length of a thread which will wind spirally about a cylinder of 4 feet in circumference, and 60 feet in length, the distance between each turn of the thread being 1 foot? Ans. 247'38, &c. feet. 33. If A can perform a piece of work in 10 days, B in 12 days, C in 16 days, then how many days will be required for all together to perform the work? Ans. 4-. 34. A shepherd, in the time of war, fell in with a party of soldiers, who plundered him of half his flock, and half a sheep over; afterwards a second party met him, who took off half of what he had left, and half a sheep over; and, soon after this, a third party met him, and used him in the same manner, and then he had only five sheep left. It is required to find what number of sheep he had at first. Ans. 47 sheep. 276 I G IER A l'1ITLMTI( C. 35. Four persons, A, B, C, and D, spent 20 shililngs in company, when A proposed to pay I, B -:, C 5, and D i part; but when the money canme to be collected, they found it was not sufficicnt to answer the intended purpose. The question then is, to find how much each person must contribute to make up tile whole reckoning, supposing their several shares to be to each other in the proportion above specified. [ A must contribute 2 0s 1 2~S, Ans.'B C ". -s. gLD'; Os aS. 36. A stationer sold quills at 10 s. 6 d, a thousand, by which he cleared - of this money; but, growing scarce, he raised them to 12s. a thousand, What did he clear per cent. by the latter price? Ans. 713 per cent. 37. How-much can a person give for an annuity of $400, which has to run 12 years, if the interest be reckoned at 3 per cent.? Ans. $3981'602. 38. A debt, due at this present time, amounting to $1200, is to be discharged in seven yearly and equal payments. What is the amount of one of these payments, if the interest be calculated at 4 per cent.? Ans. $199'931. 39. An usurer lent a person $600, and drew up for the amount a bond of $800, payable in 3 years, bearing no interest. What did he take per cent., if compound interest be taken into consideration? Ans. 10'06424 per cent. 40. A debtor, being unable to pay his debts, amounting to $12950, at once, agrees with his creditors to discharge this sum by monthly instalments, vi z,: $600 tie 1st month,I MISCEI lLA NEO U U ESTIONS. 277 and each succeeding month $50 more than the preceding one. In how many months will he have discharged his whole debts, and how much does he pay the last month? Ans. In 14 months, and $1250. 41. A person dying, leaves half of his property to his wife, one sixth to each of two daughters, one twelfth to a servant, and the remaining $600 to the poor. What was the armount of his property? Ans. $7200. 42. We know, from natural philosophy, that any body which falls in vacuo, passes, in the first second, through a space of 16~-'1 feet; and, in each succeeding second, 321 feet more than in the one immediately preceding.: Now, if a body has been falling 20 seconds, how many feet will it have fallen the last second, and how many in the whole time? Ans. 627~ feet, and 6433~ feet. 43. An estate of $7500 is to be divided between a widow, two sons, and three daughters, so that each son shall receive twice as much as each daughter, and the widow herself $500 more than all the children. What was her share, and what the share of each child? rWidow's share,. $4000. Ans. i Each son's,.... $1000. LEach daughter's,. $ 500. 44. Three soldiers, in a battle, made $96 booty, which they wished to share equally. In order to do this, A, who made the most, gave B and C as much as they already had; in the same manner, B then divided with A and C, and, after this, C with A and B. If, then, by these means, the intended equal division is effected, how much booty did each soldier make? Ans. A, $52; B, $28; and C, $16. 45. Two carpenters, twenty-four journeymen, and 24 27S 1IR1 1 1 E K1 A R 1 IITI C. eight apprentices, received, at the end of a certain time, $144. The carpenters received $1 per day, each journleyman half a dollar, and each apprentice 25 cents. How many days were they employed? Ans. 9 days. 46. A man, to please his children, brings home a number of apples, and divides them as follows: To the first and eMest of his children, he gives the half of the whole number, less 8; to the second, the half of the remainder, again diminished by 8, and he does the same with the third and fourth. After this, he gives the 20 remaining apples to the fifth. How many apples did he bring home? Ans. 80. 47. A farmer being asked how many sheep he had, answered that he had them in five fields; in the first he had 4, in the second a, in the third 8, in the fourth l'~, and in the Aftll 450. How many had he? Ans. 1200. 48. I once had an untold sumn of money lying before me. From this, I first took'away the 3d part, and put in its stead $50. A short time after, I took from the sum, thus augmented, the 4th part, and put again in its stead, $70. I then counted my money, and found $120. What was the original sum? Ans. $25. 49. After paying away I and 5 of my money, I had $66 left in my purse. How many dollars were in it at first? Ans. $120. 50. A countryman brings his eggs to market, and first sells 4 more than the half of them; then he goes further, and sells half of the remainder, and 2 over. Now 6 eggs more than half of the remainder are stolen from him, and, dissatisfied about this loss, he returned to his village with the 2 eggs which remained in his basket How many eggs did he take to town? Ans. 80 eggs. MISCELLANEOtUS QUEStIONS. 279 51; A person goes to a tavern with a certain sum of money in his pocket, where he spends 2 shillings; he then borrows as much money as he had left, and, going to another tavern, he there spends 2 shillings also; then borrowing again as much money as was left, he went to a third tavern, where, likewise, he spent 2 shillings, and borrowed as much as he had left; and, again spending 2 shillings at a fourth tavern, he then had nothing remaining. What had he at first? Ans. 3 s. 9 d. 52. A cistern can be filled by 3 pipes; by the first in 1 hours, by the second in 33 hours, and by the third in 5 hours. In what time will this cistern be filled when all 3 pipes are open at once? Ans. In 48 minutes. 53. Into what parts must 36 be divided, so that ~ of the first, I of the second, and I of the third, may be all equal to each other? Ans. 8, 12, and 16. 54. A dog pursues a hare. Before the dog started, the hare had made 50 paces, and this is the distance between them at first. The hare takes 6 paces to the dog's 5, and 9 of the hare's paces are equal to 7 of the dog's. How many paces can the hare take before the dog overtakes her? Ans. 700 paces. 55. A person wishes to dispose of his horse by lottery. If he sells the tickets at $2 each, he will lose $30 on his horse; but if he sells them at $3 each, he will receive $30 more than his horse cost him. What was the number of tickets, and what did the horse cost him? ( The number of tickets was 60. Ans. The horse cost him $150. 56. A person had two barrels, and a certain quantity of wine in each. In order to have an equal quantity in each, he poured out as much of the first cask into the 280 HIGHER ARITHMETIC. second as it already contained; then again he poured out as much of the second into the first as it then contained; and, lastly, he poured out again as much from the first into the second as there was still remaining in it. At last, he had 16 gallons of wine in each cask. How many gallons did they contain originally? Ans. The first, 22; the second, 10 gallons. 57. What is the sum of the cubes of the seven following fractions: 4 2 4 97 54 5 5 5 3, and 5 2? Ans. 6. 58. A gentleman hires a servant, and promises him, for the first year, only $60 in wages; but, for each following year, $4 more than for the preceding. How much will the servant receive for the 17th year of his engagement, and how much for all 17 years together? Ans. $124, and $1564. 59. A general, wishing to draw up his regiment into a square, tried it in two ways. The first time he had 39 men over; the second time, having extended the side of the square by one man, he wanted 50 men to complete the square. What was the number of soldiers in the regiment? Ans. 1975 soldiers. 60. From a sum of money, $50 more than the half of it is first taken away; from the remainder, $30 more than its fifth part; and again, from the second remainder, $20 more than its fourth part. At last there remained only $10. What was the original sum? Ans. $275. 61. HIERo, king of Sicily, ordered a crown to be made, containing 63 oz. of pure gold; but, suspecting that the workmen had debased it by using part silver, he recommended the detection of the fraud to the famous ARCHIMEDES, who, putting it into water, found that it MiSCEPiLANLOf0US Q UESTIONS. 281 displaced 8-2245 cub)ic inches of water. He next found that-a cubic inch of gold weighed 10'36 ounces, and a cubic inch of silver weighed 5'85 ounces; then, from this data, he calculated the proportions of gold and silver of which the crown was composed. What must have been his result?'n 25447 oz. of gold to every 2'1434 oz. of silver. 62. The number 48 is divided into two parts, such that if the greater part be multiplied by 24, and the less part by 12, the difference of their products will be 504. What are the parts? Ans. 30, and 18. 63. A man and his son together worked 50 days. The man received $1 per day, and the son $0'75. Their whole wages amoulnted to 844. How many days did each work? A Ts' he man worked 26 days., The son " 24 " 64. Three men, A, B, and C, together work 60 days, and all receive $48; A received $0'70 per day, B $0'80, and C $0'90. How many days did each work? [ A and C worked each the same I number of days, and B worked Ans. the balance of the 60 days, be l the same more or less. 65. Two men, A and B, are on a straight road, on the opposite sides of a gate, and distant from it 308 yards and 277 yards, respectively, and travel each towards the original station of the other. How long must they walk so that they shall be equally distant from the gate, if B goes 2 yards per second, and A 21 yards per second? Ans. 1 minute, 33 seconds. 66. Every thing being supposed as in the preceding 24 * t282 HIGHER ARITIIMETIC. question, at what time will each be at tile saute distance from the original station of the other as the other is' from his? Ans. 4'- minutes after starting. 67. Suppose the elastic power of a ball which falls from a height of 100 feet, to be such as to cause it to rise 50 feet, or one half tile height from which it fell, and to continue in this way, diminishing the height to which it will rise, in geometrical progression, till it come to a state of rest, how far will it have moved? Ans. 300 feet. 68. What is the square of 12890625? Ans. 166168212S90625. NOTE.-In this question, it will be observed that the square of the above number ends with the same set of figures as the number itself; and this must hold good for any power of the above number. The same is true of the number 109376. Either of these numbers can be carried to any extent, by prefixing the proper figures. These are the only numbers which possess this remarkable property. APPENDIX. IN this Appendix we shall, by using the Roman notation, number the different portions the same as the Articles in the body of the work, which they are designed to explain or enlarge upon. II. The particular form of many of the symbols employed to indicate the relations of numbers, as well as the different operations which are required to be performed upon them, are no doubt entirely arbitrary. But some of them are easily traced to their true origin, and a reason is readily seen why they were employed. Under this article we propose to make a few remarks on what may be properly called the philosophy of arithmetical symbols. - Not that we presume to know positively the reasons for the particular form of all symbols, but it is thought a few suggestions and queries on this subject might not be uninteresting to the inquiring mind. The most natural method which could be devised for indicating that one quantity or number was greater than another is the symbol >, or <, so placed between the quantities or numbers compared as to open towards the greater, showing that in passing from the angle of this symbol to the opposite side, there is an increase. In this sense, this same symbol is used in music. In conformity with this use of the above symbol we have $1>99 cents; 45 cents<$1; &c. The symbol $, employed to denote dollars is no doubt derived from a 284 APPEND1)X. combination of the letters U and S, which, taken together, form the abbreviation for United States. If a compound letter be formed by uniting these letters, it will at once be seen how small a modification it is necessary to make in it so as to obtain the symbol $. And when we consider that this kind of currency, having a dollar for its unit, is peculiarly that of the United States, the explanation seems quite satisfactory. Having agreed to employ the symbol >, or <, to denote inequality, we may ask how must it be modified when placed between two equal quantities so as to become the symbol of equality? In this case, there is no reason why this symbol should be placed with its openings towards the one quantity rather than the other, since neither is larger than the other. It ought to open equally towards each; hence it must of- necessity become =, which is the symbol of equality. The addition of one quantity to another might very concisely be represented by the union of two straight lines. But in what manner shall these two lines be united? When numbers are united by addition, the first may be said to be added to the second, or the second may be said to be added to the first, so that the two terms united have mutual relations with each other. So also when two numbers are united by multiplication, they are each factors, and have mutual relations. And moreover since the operations of multiplication are so nearly allied to those of addition, it would seem natural to represent multiplication also by the union of two lines so placed as to be mutually related. If two equal lines mutually bisect each other at right angles, it would form a symbol which might be appropriately used to denote APPENDIX. 285 either addition or multiplication. When these lines are so placed as that one may be horizontal and the other perpendicular, as +, it is the symbol of addition. When this symbol is so placed that the lines are oblique as x, the symbol denotes multiplication. Since the symbol +, denotes the union of two numbers by addition, if we take one away, or perform the operation of subtraction, this operation might be indicated by the symbol + after one of the lines, as for instance the vertical one, had been taken away, so as to give - for the symbol of subtraction. The symbol -, for division, was probably suggested by the idea of separating or dividing by means of a line passing through the quantity, leaving a portion on each side of the straight line, which portions are represented by the two dots placed on the two sides of this straight line. If we substitute numbers for those dots, considering the one above the line to correspond with the dividend, and the one below the line to correspond with the divisor, we shall thus have, in the case of 7 divided by 5, the expression 7, which is in strict accordance with the notation for a vulgar fraction, which shows that a vulgar fraction is but another method of denoting a division where the numerator corresponds with the dividend, an'd the denominator with the divisor. If in the proportion 6: 3 8: 4, we draw a straight horizontal line between the two dots separating the first and second terms, as well as between the two dots which separate the third and fourth terms; also draw two horizontal lines joining the four dots between the second and third terms, we shall have 6-3:=: 84, 286 APPENDI)IX. which is read, 6 divided by 3 is equal to S divided by 4, and this is precisely what the first expression, 6: 3:: 8: 4, meant. The symbol