176. Finding the G.C.M. of Large Numbers. It is not often necessary to find the G.C.M. of large numbers that cannot be separated readily into factors. An example will illustrate a process often called the long-division method.

This method depends upon the principle that a common factor of two numbers is also a factor of their sum and of their difference.

Required the G.C.M. of 323 and 357.

323)357(1

323

34)323(9
306
17)34(2
34

Here we cannot easily factor the two numbers. We divide the larger number by the smaller to see if the smaller is the G.C.M. It is not because there is a remainder 34. We then divide the smaller number by this remainder and find there is a remainder 17. Hence 34 is not the G. C.M. Then we divide the last divisor 34 by the last remainder 17 and find that 17 is an exact divisor of 34. Then 17 is the G. C. M. required.

17. What is the largest number that will exactly divide 2378 and 3915?

18. What is the largest number that will exactly divide 551, 667, and 841 ?

177. Multiple. The product of two or more integers is called a multiple of each of them.

Thus 20 is a multiple of 2, 4, 5, and 10, because 2 × 2 × 5 = 4 × 5 = 2 × 10 = 20. This has already been defined (§ 36). We speak of 75, for example, as a multiple of 25, and of 51 as a multiple of 17, or we speak of each as a multiple of any other of its factors.

178. Common Multiple. A multiple of two or more numbers is called a common multiple of the numbers.

Thus 60 is a common multiple of 6 and 10, and so are 30, 90, 120, and so on, common multiples of 6 and 10.

179. Least Common Multiple. The smallest common multiple of two or more numbers is called their least common multiple. It is the least number that is exactly divisible by each of them.

Thus 60 is a common multiple of 6 and 10, but the least common multiple is 30.

The letters L. C. M. are used for the words Least Common Multiple. Since the L.C. M. must contain each of the numbers, it must contain all of the prime factors of each, and this is the principle upon which the finding of the L. C. M. rests.

180. Finding the L.C.M. Required the L.C.M. of 24, 30, 36. Resolving each number into its prime factors, we have

The L.C.M. must contain the factors of 24, which are 2, 2, 2, and 3. These are therefore taken as part of its factors.

It must contain the factors of 30, which are 2, 3, and 5; but we have already taken 2 and 3, so we need take only 5.

but we

It must contain the factors of 36, which are 2, 2, 3, and 3; have already taken 2, 2, and 3, so we need take only one more 3. Hence the factors of the L. C. M. are 2, 2, 2, 3, 5, 3. Their product is 360, and 360 is the L.C.M.

It is often convenient to adopt another arrangement of the work, as in finding the L.C.M. of 12, 18, 30, 45.

Here we arrange the numbers in a horizontal line, and divide by the smallest prime factor common to two or more of them.

2)12 18 30 3 6 9 15

2

45 45 15

We first divide by 2, and write the quotients and the undivided numbers as well in a line below. In the first line of quotients we cancel the 9 and the 15 because each is a factor of 45, and therefore 45 contains all the We next divide by 3, and in the last line of quotients we have the numbers 2 and 15, which are prime to each other. Hence the L. C.M. is 2 x 3 x 2 x 15, or 180.

factors of 9 and of 15.

Required the L.C.M. of 5, 15, 30, 60.

5 13 30 60

Here we cancel the 5 because it is a factor of 15. We cancel the 15 because it is a factor of 30, and cancel the 30 because it is a factor of 60. Therefore 60 contains all the factors of the other numbers and is itself the L.C.M.

181. General Directions for Finding the L.C.M. From the above examples the following directions may be observed: Separate each number into its prime factors in any convenient manner. Find the product of these factors, using each factor the greatest number of times it occurs in any one of the given numbers.

39. Find the least number of marbles that can be divided equally among 15, 20, or 25 boys.

40. Find the shortest distance that can be exactly measured by a 3-foot rule, or a 5-foot pole, or a 10-foot pole.

41. What is the smallest quantity of liquid that can be measured exactly by 2-quart, 4-quart, 6-quart, or 8-quart measures?

42. What is the smallest sum of money that can be counted out in 5-cent pieces, 10-cent pieces, 25-cent pieces, or 50-cent pieces?

43. What is the capacity of the smallest water tank that can be filled in an integral number of minutes by either of two pipes, if one discharges 32 gallons a minute and the other 48 gallons a minute?

44. Find the least number of oranges that, when arranged in groups of 6, 7, 8, or 9, has just 5 over in each case. (Find the L.C.M. and add 5. Test the result.)

45. Find the least number of marbles that, when arranged in groups of 8, 12, 15, 60, or 120, has just 7 over in each

case.

182. Cancellation. As in § 97, we may apply our knowledge of divisibility of numbers, and shorten the work by canceling equal factors from both dividend and divisor.

For example, divide 11 × 27 × 30 by 3 × 9 × 15.

The common factor 3 is canceled from the 3 in the divisor and the 27 in the dividend. The common factor 9 is canceled from the 9 in the divisor and the 9 that remains uncanceled in the 27 of the dividend.

1 × 1 × 1. Hence the quotient is 22. This is much shorter than finding two products, 8910 and 405, and then the quotient 22.