7. To Find the L.C.M. of 21 and 35. The factors of 35 are 5 and since 3 and 5 are prime to each other, therefore the common factor 7 is the G.C.D. of 21 and 35. Now the number required cannot be less than 21, because it must contain 21; and if it contains 21, it must contain all the prime factors of 21, namely, 3 and 7. Again, the number must also contain 35, and therefore it must contain the prime factors of 35, namely, 5 and 7. Therefore the number 3 × 7 × 5 which contains all the prime factors of 21 and 35, and no others, is the L.Č.M. But (217) × 35 = 3 × (5 × 7) = 105, and 21x (357)=(3x7) x 5 = 105. 48. Hence the following rule for finding the L.C.M. of two numbers : Rule.-Divide either of the numbers by their G.C.D., and multiply the quotient by the other number. 49. If the numbers are prime to each, the L.C.M. is their product. 50. To find the L.C.M. of three or more numbers, we may find the L.C.M. of two of the numbers, then the L.C.M. of that multiple and a third number, and so on. The multiple last found will be the L.C. M. of all the numbers. 51. This rule should always be used when the prime factors of the numbers cannot be found by inspection; but when this can be done the result may be more readily obtained by either of the following methods: 52. First Method. Rule.-Strike out every number which is con tained in any other number; and separate each of the remaining numbers into its prime factors. Then, if all the prime factors are different, the L.C.M. will be found by multiplying them all together; but if the same factor appears in two or more of the given numbers it must be repeated the greatest number of times which it appears in any of the given numbers. Examples. (1) Find the L.C.M. of 14, 21, 15, 25, 30, 9. 14 = 2 × 7 21=3×7 25=5×5 Explanation.-Strike out 15, because it is contained in 30, and separate each of the other numbers into its prime factors. The 30= = 2 × 3 × 5 number required must contain the 9=3×3 factors of 14, namely, 2x7; it must also contain the factors of 21, namely, 3 and 7; but of these the 3 only must be set down, because the factor 7 has been already taken. Thus the number 2 × 7 × 3 contains all the prime factors of 14 and 21, but the number required must contain the factors of 25, namely, 5 and 5; annexing these to the series of factors, we get and since this number contains 2, 3, 5, which are all the prime factors of 30, no additional factors must be set down for 30. Lastly, the number required must contain the factors of 9, namely, 3 and 3; annexing one of these factors we get the number 2 × 7 × 3 × 5 × 5 × 3 which contains all of the prime factors of the given numbers, and no others, and therefore 2 × 7 × 3 × 5 × 5 × 3 or 3150 is the L.C.M. required. X 53. Second Method. Rule. Write down all the numbers in a line, with a comma between every two. Divide by any prime factor which will divide two or more of the given numbers, and write the quotients and undivided numbers in a line underneath. In like manner divide the quotients and undivided numbers, and continue the process until a row of numbers is obtained which are prime to each other. The product of all the divisors and the numbers in the last line is the L.C.M. required. Find the L.C.M. of 14, 21, 15, 25, 30, 9. Collecting the divisors and the numbers in the last line we get 2 × 7 × 3 × 5 × 5 × 3, or 3150 for the L.C.M., as before. The operation may be shortened by omitting the I's and striking out in each line any number which is contained in any other number in the same line. Thus Examples.-1. Find the L.C.M. of 1, 2, 3, 4, 5, 6, 7, 8, 9. .. The L.C.M. required = 5 × 2 × 3 × 7 × 2 × 2 × 3 =2520. Second Method. 2)1, 2, 3, 4, 5, 6, 7, 8, 9 5, 3, 7, 4, 9 ... The L.C.M. required = 2 × 5 × 7 × 4×9 = - 2520. (2) Find the L.C.M. of 24, 36, 18, 15, 25, 20. 24, 36, 18, 15, 25, 20 24 = 2 × 2 × 2 × 3 15=3×5 25=5×5 20 = 2 × 2 × 5 X .•. L.C.M. required = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800. ... L.C.M. required = 2 × 2 × 3 × 2 × 3 × 25 = 1800. EXERCISE 10. Least Common Multiple. Find the L.C.M. of A. (1) 8, 12, 9; (2) 14, 21, 15; (3) 18, 24, 25; (4) 42, 28, 35; (5) 16, 25, 40; (6) 48, 36, 54; (7) 63, 84, 108; (8) 50, 28, 70; (9) 72, 96, 45; (10) 81, 32, 125. B. (1) 2, 3, 4, 6, 8, 9, 10; (2) 3, 4, 5, 6, 7, 8, 9; (3) 8, 10, 6, 14, 21, 25, 18; (4) 12, 15, 18, 16, 20, 24, 30; (5) 9, 14, 21, 28, 42, 45, 54; (6) 7, 26, 35, 25, 28, 52, 65; (7) 56, 84, 91, 75, 63, 112, 80; (8) 76, 57, 96, 95, 114, 125, 160; (9) 27, 36, 81, 144, 100, 105, 56; (10) 18, 20, 24, 25, 27, 132, 165. C. (1) 289 and 323; (2) 667 and 899; (3) 403 and 527; (4) 731 and 493; (5) 1081 and 851; (6) 779 and 1007; (7) 372, 192, and 156; (8) 306, 204, and 136; (9) 364, 273, and 143; (10) 294, 245, and 175. Separate into their prime factors, and thence find the G.C.D. and L.C.M. of— D. (1) 12, 8, 20, 16; (2) 18, 24, 54, 12; (3) 36, 27, 45, 18; (4) 21, 14, 35, 28; (5) 40, 24, 16, 56; (6) 25, 20, 15, 35; (7) 36, 60, 24, 72; (8) 3c, 48, 18, 42; (9) 120, 84, 96, 36; (10) 132, 72, 126, 84. CHAPTER X. VULGAR FRACTIONS. 54. An integral number or integer is either a unit, or a collection of units or wholes. 3, 4, 17 are integral numbers or integers. Thus, I, M |