tiple. 240; therefore it is the least common multiple of all these numbers. Or the above principle may be illustrated as in the following question. 2. What is the least common multiple of 4, 5, 6, 7, 8, and 9 ? 2)4 5 6 7 8 9 2)2 5 3 7 4 9 In this question, we divide by the in 2)1 5 3 2 9 prime factors, 2, 3, 5, 7, &c., succes3)1 5 3 7 1 9 sively, until the last quotients terminate 3)1 5 1 1 3 in units; and the product of all the 51 5 1 1 i prime factors is the least common mul. 7)1 1 1 0 1 1 1 1 1 1 1 1 Thus 2 x 2 X2 X3 X3 X 5x7 = 2520 Ans. That 2520 is the least common multiple is evident from the fact, that the divisors whose product produces this number are the prime factors, and the only prime factors of the given numbers, as may be seen below. Factors of 4 are 2 and 2 = 2 X 2=4 5 and 1=5 X 7 and 1=7 x1=7 3 and 3=3x3=9 RULE. — Divide by such a number as will divide most of the given numbers without a remainder, and set the several quotients, with the several undivided numbers, in a line beneath, and so continue to divide, until no number greater than unity will divide two or more of them. Then multiply all the divisors, the last quotients, and undivided numbers together, and the product is the least common multiple. Or, divide the given numbers successively by their prime factors, until the last quotients terminate in units. The continued product of all the divisors will be the least common multiple. 3. What is the least common multiple of 6, 8, 10, 18, 20, and 24 ? Ans. 360. 4. What is the least common multiple of 14, 19, 38, and 57? Ans. 798. 5. What is the least common multiple of 20, 36, 48, and 50 ? Ans. 3600. 6. What is the least common multiple of 15, 25, 35, 45, and 100 ? Ans. 6300. 7. What is the least common multiple of 100, 200, 300, 400, and 575 ? Ans. 27600. 8. I have four different measures; the first contains 4 quarts, the second 6 quarts, the third 10 quarts, and the fourth 12 quarts. How large is a vessel, that may be filled by each one of these, taken any number of times full ? Ans. 60 quarts. Note. - In finding the common multiple of two or more numbers, any one number that can measure another may be cancelled. 9. What is the least common multiple of 4, 6, 8, 12, 16, 10, and 20 ? 44 6 8 12 16 10 20 4x3x4x5=240 Ans. 3 4 5 By examining this question, we find that 8 may be divided by 4, 12 by 6, 16 by 8, and 20 by 10; therefore we cancel 4, 6, 8, and 10. 10. What is the least common multiple of 5, 15, 30, 7, 14, and 28 ? 2)$ 16 30 7 14 28 2X15X14=420 Ans. 15 14 In this question, we find that 15 may be measured by 5, 30 by 15, 14 by 7, and 28 by 14; we therefore cancel 5, 15, 7, and 14. 11. What is the least common multiple of 1, 2, 3, 4, 5, 6, 7, 8, and 9? 2)1 % 8 4 5 6 7 8 9 2X3X5X7X4X3=2520 3)5 3 4.9 [Ans. 5 17 4 3 12. What is the least common multiple of 9, 8, 12, 18, 24, 36, and 72 ? Ø 8 12 18 24 36 72 72 Ans. 13. What is the least number that 18, 24, 36, 12, 6, 20, and 48 will measure? To reduce fractions to a common denominator, that is, to change fractions to other fractions, all having their denominators alike, yet retaining the same value. 1. Reduce , and to other fractions of equal value, having the same, or a common, denominator. First Method. OPERATION. 4)8 12 16 4X2X3X2-48 common denominator. 2)2 3 4 81 6X 7=42 numerator for } =*. 1 3 2 12 4X 5=20 numerator for =o. 16/ 3X11=33 numerator for if= Having first obtained the least common multiple of all the denominators of the given fractions by the last rule, we assume this as the common denominator required. This number (48) we divide by the denominators of the given fractions, 8, 12, and 16, and find their quotients to be 6, 4, and 3, which we place under the 48; these numbers we then multiply by the numerators 7, 5, and 11, and find their products to be 42, 20, and 33, and these numbers are the numerators of the fractions required. Second Method. OPERATION. 11 X 8 X 12 = 1056 numerator for H = Home Tho numerators are produced by multiplying the numerators of the given fraction by each of the other denominators, and the common denominator is obtained by multiplying all the denominators. By this process, we obtain the following Ans. 1344, 46, 1956. The pupil will perceive, that this method does not express the fractions in so low terms as the other; although they both have the same value. 1344 RULE. — Find the least common multiple of all the denominators by Case VII., and it will be the denominator required. Divide the common multiple by each of the denominators, and multiply the quotients by the respective numerators of the fractions and their products will be the numerators required. Or, multiply each numerator into all the denominators except its own for a new numerator ; and all the denominators into each other for a common denominator. Questions to be performed by the first method. 2. Reduce, , }, and is Ans. 4, 48, ,9. 3. Reduce 1, 185, 18, and 24. Ans. 773327,132, 116, 4. Reduce 1, 14, 2d, and so Ans. 34, 1, 3, t. 5. Reduce ži, , 11, and t. Ans. 3, 1, 1, 1. 6. Reduce , 1, 4, and . Ans. 318, 304, 175, 331. 7. Reduce 3, , , and Ans. 8, 48, 6, . 8. Reduce , s, t, and 11: Ans. 34, 8, 11, 44. 9. Reduce , Š, ], and 3. Ans. , 135, og til. 10. Reduce , 3, 4, and 4. Ans. 175 11. Reduce , 1, 1, and ]. Ans. 41, 42, 43, 41. 12. Reduce $, iz, 13, and . Ans. 231, 233, 23, 13. Reduce 18, 30, 31, and 36. Ans. 78, 40, 66, 60%. 14. Reduce 3, tai to, and Jo. Ans. 218, 148, 16, 15. Reduce ş, 7, 8, and 54. Ans. 44, 4 , 3**. Questions to be performed by the second method. 16. Reduce, , and to fractions having a common denominator. Ans. So, 198. · 17. Reduce 4, , and for Ans. 3966, 8, 1918. 18. Reduce , #, and . Ans. T, Bota Sot 19. Reduce 7o, , and 7. Ans. 44, 123, . 20. Reduce 17, 5, and to 21. Reduce , ty, and 1115. Ans. 2995 205, 2015 22. Reduce }, , 4, and 8. Ans. 42, 43, 44, 46. 23. Reduce , Ýr, and of 75. Ans. , 750, 64326. 24. Reduce 3, 5, , and 17. Ans. , . 25. Reduce 135 of 6, and 213. Ans. 118, 126, 438. 26. Reduce q, tr, 4, and J. 47 2 5 10 20 6 1 2 298, 2293 2295 4 27. Reduce 175, 117, and 1725 1993, Note. 1. - If there be complex fractions, they must first be changed to simple fractions, and then they may be reduced by the foregoing Rule. 28. Reduce a common denominator. Ans. 54, , 38. to a common denominator. Ans. 36, 38, 30. Reduce 16, , and of 8to a common denominator. Ans. 4935, 40909.. Note 2.- To reduce complex fractions to equivalent fractions, having a given fraction for a common denominator, we first change the complex fractions to simple fractions, and multiply each of their numerators by the numerator of the required common denominator, and their denominators by the denominator of the required denominator ; under these fractions we write the required denominator. 31. Change and to other fractions of equal value, having for a common denominator. OPERATION. =f; X = =}, first numerator. CASE IX. This question may be thus anabox f. = =bs. lyzed. Since 12 pence make a shil27 X 35. = 610£. ling, there will be as many shil lings as pence; therefore i of of a penny is it = of a shilling. Again, as 20 shillings make a pound, there will be to as many pounds as shillings; there. fore o of " of a shilling is oo of a pound. Q. E. D. This question may be abridged. Thus, i X 1 X 2+ = Toko = odo Ans. Rule. — Multiply the denominator of the given fraction by all the denominations between it and the one to which it is to be reduced, and over the product write the given numerator. 2. Reduce of a farthing to the fraction of a pound. and I Ans. |