12 X 20 I2 X 20 7 20 so that 12 x 20 20 2 20 by 15, is not a multiple of 20. If we next try 6 for divisor, we shall find that is a multiple of 20, but not of 12; 6 being 6 a measure of 12, but not of 20. For, 6 being a measure of 12, is equivalent to a whole number (2); so that which 6 6 is the product of 20 by 12, is a multiple of 20. . But as 6 is not 6 a measure of 20, 5 is not equivalent to a whole number ; which is the product of 12 by is not a multiple of 12. 6 6' Let us now try 2 for divisor, and we shall find that is a common multiple of 12 and 20; 2 being a common measure. For, as 2 measures 20, is equivalent to a whole nunıber (10); which is the product of 12 by 22, is , is a multiple of 12. Again : as 2 measures 12, is equivalent to a whole 12 number (6); so that which is the product of 20 by 2' is a multiple of 20. It thus appears that the divisor must be a common measure ; and upon the principle that—the dividend remaining the same—the quotient is smallest when the divisor is largest, we obtain the least common multiple of 12 and 20 when we divide 12 x 20 by 4, the greatest common measure. General demonstration.—Let x and y be any two numbers, and m their greatest common measure: to prove that -is the least common multiple of x and y. As m is a measure of y, Y is equivalent to a whole number; and xx y xy is a multi. 2 I2 X 20 20 so that 2 2 12 X 20 2 ple of X. Again: as m is a measure of x, is equivalent to a XY 2 or m whole number; and yx is a multiple of y. So that is a common multiple of x and y. It is also the least common multiple. For, if not, it must (p. ху 112) be a multiple of the least. Let a times the least common multiple, if possible : in other words, let the least common I multiple be (=a=) Then, because on which is the am am' xy XY , m y product of x by is a multiple of 2, Y must be equivalent am to a whole number: so that am must measure y. And be xy which is the product of y by-, is a multiple of y, am' am cause am m must be equivalent to a whole number : so that am am must measure x. It thus appears that am is a common measure of x and y. But this is impossible, the greatest common measure of and y being m. It follows, therefore, that ay is the least common multiple of x and y. 101. To find the Least Common Multiple of more than two numbers : Find, first, the least common multiple of two of the numbers; next, the least common multiple of this common multiple and a third number; then, the least common multiple of this new common multiple and a fourth number; and so on, until all the numbers have been taken into account. The last common multiple will be the least common multiple required. and 27. 72 216 EXAMPLE.—Find the least common multiple of 8, 12, 18, The least common multiple of 8 and 12 is 8 12 18 27 24; of 24 and 18, 72; and of 72 and 27, 216. So that, according to the Rule, 216 is 24 the least common multiple required. Reason of the Rule.-As 24 is a common multiple of 8 and 12, 8 and 12 are measures of 24, and therefore of 72-a multiple of 24; 18 also is a measure of 72; consequently 8, 12, and 18, being measures of 72, are measures of 216–a multiple of 72. In other words, 216 is a common multiple of 8, 12, and 18: therefore, being a multiple of 27 also, 216 is a common multiple of the four given numbers. It is also the least common multiple. For, if a smaller number than 216 were a common multiple of 8, 12, 18, and 27, this smaller number, being a multiple of 8 and of 12, would (p. 112) be a multiple of 24—the least common multiple of 8 and 12; and, being a multiple of 24 and of 18, would be a multiple of 72— 2 a с a a the least common multiple of 24 and 18; and, being a multiple of 72 and of 27, would be a multiple of 216—the least common multiple of 72 and 27. But a smaller number than 216 cannot be a multiple of 216: neither, therefore, can a smaller number than 216 be a common multiple of 8, 12, 18, and 27. So that 216, which has been shown to be a common multiple, is the least common multiple. General Demonstration.—Let w, x, y, y and z be any four numbers, and let the least common multiple of w and x be a ; of a and y, b; and of b and z, c: to prove that c is the least common multiple of w, X, Y, and z. Because w and x measure a, they measure b—a multiple of a; y also measures b; consequently w, x, and y, being measures of b, measure c-a multiple of b. In other words, c is a common multiple of w, x, and y: therefore, being a multiple of z also, c is a common multiple of the four given numbers. It is also the least common multiple. For, if possible, let a smaller number than c be a common multiple of w, x, y, and z. Then, this smaller number, being a common multiple of w and x, is a multiple of a—the least common multiple of w and x; and, being a common multiple of a and y, is a multiple of b— the least common multiple of a and y; and, being a common multiple of b and z, is a multiple of c—the least common multiple of b and 2. But a smaller number than c cannot be a multiple of c: neither, therefore, can a smaller number than c be a common multiple of w, x, y, and z. Consequently c, which has been proved to be a common multiple, is the least common multiple. 102. Another Rule for finding the Least Common Multiple of more than two numbers : Arrange the numbers in a horizontal row, and divide as many as possible by a common measure. Set down, in a second horizontal row, the resulting quotients and the undivided numbers. Treat this and every succeeding row in the same way as the first, and continue the process until a row of numbers shall have been obtained, of which every two are prime to each other. Then, multiply the several divisors and the numbers in the last row together, and the product will be the least common multiple required. IO 3) I I I EXAMPLE.–Find the least common multiple of 50, 60, 80, 90, and 100. These numbers are all measured by 10. Dividing by 10, we obtain for quotients 5, 6, 8, 9, and 10. Dividing as many as possible 10) 50 60 80 90 100 (i.e., 6, 8, and 10) of these quo- 2) 5 6 8 9 tients by a common measure,, 2, 5) 5 3 4 9 5 and bringing down the undivided 3 numbers, we obtain 5, 3, 4, 9, and 4 9 5. Of this row of numbers, we can 4 3 divide not more than two (the first and the last, or the second and the fourth) by a common measure. Dividing the first and the last by 5, and bringing down the undivided numbers, we obtain 1, 3, 4, 9, and 1. Dividing the second and the fourth of this last row of numbers by 3, and bringing down the undivided number 4,—it is obviously unne. cessary to bring down the ones,—we obtain 1, 4, and 3: of which numbers every two are prime to each other. So that, according to the Rule, the least common multiple required is 10X2X5X3X4X3=3,600. Reason of the Rule.-Upon the principle that (when there is no remainder) a dividend is equal to the product of the divisor by the quotient, we have 50 60 I0X2X5X3 1oxzx;} 10X6= 10X2X5X3X4 10X3X3 10X2X5X3X4X3 I0X10= 10X2X5. 10X2X5X3X4X3 Now, the least common multiple of the first two numbers (10X 5 and 10x2x3)—the greatest common measure being 10- is (10x5) X(10x2x3) = 10=10X2 X 5 X 3. The least common multiple of this common multiple and the third number (10X2 X 4)—the greatest common measure being 10X2is (10 x 2X5 X 3) x (10 X 2 X 4) = 10 X2=10X2X5X3X4. The least common multiple of this common multiple and the fourth number (10x3x3)— the greatest common measure being 10x3-is (10 X 2 X 5 X 3 X 4) (10x3x3): 10 x 3 = 10 X 2x5*3*4*3. And the least common multiple of this last common multiple and the fifth number (10x2 x 5)—the former number being a multiple of the latter-is 10x2x5X3 X4 IOO 8 IO 4 x 3. So that (§ 101) 10 x 2 x 5x3 x4x3 is the least common multiple of 50, 60, 80, 90, and 100. Note.- When two numbers in the same row are related as measure and multiple, we save unnecessary trouble by rejecting the measure. Thus, rejecting 50, which measures 100, and dividing 10) 50 60 80 90 the remaining numbers by 10, we 2) 6 9 have in the second row 6, 8, 9, and 3 9 5 10. Dividing as many as possible (i.e., 6, 8, and 10) of the numbers in the second row by a common measure, 2, and bringing down the undivided number (9), we have in the third row 3, 4, 9, and 5. Rejecting 3, which measures 9, we find that every two of the remaining numbers are prime to one another. And we then obtain the least common multiple by multiplying 10, 2, 4, 9, and 5 together. General demonstration.—Before proceeding to examine the general demonstration given below, we must understand the following two facts : I. A number which is prime to two or more others is prime to their product. II. If the division of two numbers by a common measure give quotients which are prime to each other, that common measure is the greatest common measure of the two numbers. (I.) The first of these facts is rendered intelligible by the following considerations : (a) A (composite) number is said to be resolved into its prime factors when it is resolved into two or more factors, every one of which is a prime number. The number 30, for instance, when written 3x10, is not resolved into its prime factors: for, although 3 is, 10 is not, a prime number. But 30 is resolved into its prime factors when written 3X2X5; 3, 2, and 5 being all prime numbers. (6) If two numbers, x and y, be so related to each other that no one of the prime factors of x is a factor of y, x and y are prime to one another. For, if x and y be not prime to one another, they have some common measure, m, which must be either a prime or a composite number. Now, m cannot be a prime number : for, if it were, one of the prime factors of x would be a factor of y. Neither can m be a composite number : for, if it were, it would be resolvable into two or more prime factors, each of which would measure x and y, multiples of m; so that in this case, also, one of the prime factors of. x would be a factor of y. It thus appears that it and y, having neither a prime nor a composite number for common measure, are prime to one another. |