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OPERATION.

SEOOND METHOD. 122. This method consists in resolving the numbers into their prime factors, and taking the product of the common factors.

1. Find the greatest common divisor of 42, 84, and 126. SOLUTION.—The factors of 42 are 2, 3, and 7; the factors of 84 are 2, 2, 3, and 7; the factors of 42=2X3X7 126 are 2, 3, 3, and We see that 2, 3, and 7

8452X2X3X7 are all the prime factors common t the three

126=2X3X3X7 numbers; hence their product, which is 42, is the

2x3x7=42 greatest common divisor of the numbers (Prin. 2). Hence the following

Rule.- Resolve the numbers into their prime factors, and take the product of all the common factors.

Find the greatest common divisor of 2. 270, 315, and 405. L

Ans. 45. 3. 168, 192, and 216.

Ans. 24. 4. 252, 308, and 364.

Ans. 28. 6. 504, 546, and 588.

Ans 42. 6. 392, 448, and 504.

Ans. 56. 7. 432, 504, and 648.

Ans. 72. 8. 792, 864, 936, and 1008.

Ans. 72. 9. 384, 576, 768, and 960.

Ans, 192.

CASE II.

123. When the numbers are large and cannot be readily factored.

1. Find the greatest common divisor of 32 and 56.

SOLUTION.—We divide 56 by 32, the divisor 32 OPERATION. by 24, and the divisor 24 by the remainder 8, and have no remainder; then will 8 be the greatest

32)56(1

32 common divisor of 32 and 56. For 1st. The last remainder, 8, is a NUMBER OF TIMES

24)32(1 the G. C. D. Since 32 and 56 are each a number of

24 times the G. C. D., their difference, 24, is a number 8)24(3 of times the G. C. D., by Prin. 3; and since 24 and

24 32 are each a number of times the G. C. D., their difference, 8, is also a number of times the G. C. D.

2d. The last divisor, 8, is ONCE the G. C. D. Since 8 divides 24 it will divide 24+8, or 32, by Prin. 3; and since it divides 32 and 24, it will divide 32+ 24, or 56; and now since 8 divides 32 and 56, and is a number of times tho G. C. D., it must be once the G. C. D.

OPERATION.

ANOTHER FORM.- In the margin on the right is another form of writing the division, which in practice we prefer to the above. The problem is to find the greatest common divisor of 32 and 116.

The method will be clear from a slight inspection of the work. Let the pupils adopt it after they are familiar with the common form.

32 116/3

96 20 20/1 121 121 8 81 4 82

Rule.- Divide the greater number by the less, the divine by the remainder, and thus continue to divide the last divisor by the last remainder until there is no remainder ; the last divisor will be the greatest common divisor.

NOTE.-To find the greatest common divisor of more than two numbers, we first find the greatest common divisor of two of them, then of that divisor and one of the other numbers, etc.

WRITTEN EXERCISES. Find the greatest common divisor of 2. 115 and 161.

Ans. 23. 3. 91 and 143.

Ans. 13. 4. 333 and 592.

Ans. 37. 5. 697 and 820.

Ans. 41. 6. A farmer has two heaps of apples, one containing 364, and the other 585, which he wishes to divide into smaller heaps, each containing the same number; what is the largest number that the heaps may contain ?

Ans. 13. 7. A benevolent society distributed $678, $906, and $1146 in equal sums to the poor of three wards of a city, the sums being as large as possible. Required the amount of the equal sums and the number of persons receiving relief in each ward ?

Ans. $6; 113; 151; 191.

SUPPLEMENTARY PROBLEMS.

To be omitted unless otherwise directoa. 8. 1220 and 2013.

Ans. 61. 9. 730, 1241, and 1460.

Ans. 73. 10. 72491 and 103121.

Ans. 1021. 11. 347387 and 561851.

Ans. 1117. 12. A Western landholder has three tracts, the first containiug 583 acres, the second 574 acres, and the third 861 acres, which he wishes to divide into fields of equal size, having the least number possible. Required the number of fields and the number of acres in oach,

Ans. 48 fields; 41 aoro.

INTRODUCTION TO COMMON MULTIPLE.

MENTAL EXERCISES. 1. What number is three times 57 four times 6? five times 6? six times 87

2. A number which is one or more times another number is called & multiple of that number.

3. What is the multiple of 47 of 6? of 67 of 77 of 89 of 97 e 103 of 117 of 12?

4. What multiple is common to 2 and 3? to 3 and 49 to 4 and 67 to 8 ard 89 to 6 and 99

5. What may we call a multiple common to two or more numbers? Ans. A common multiple.

6. What is a common multiple of 4 and 57 8 and 9? 6 and 7? 4 and 69 5 and 87

7. What is the least multiple common to 2 and 4? 4 and 6? 4 and 8? 6 and 8? 8 and 12?

8. What shall we call the least multiple common to two or more Qumbers? Ans. Their least common multiple.

9. What is the least common multiple of 4 and 6? 9 and 129 10 and 16 ? 20 and 247 25 and 307

LEAST COMMON MULTIPLE.

124. A Multiple of a number is one or more times the number; thus, 4 times 5, or 20, is a multiple of 5.

125. A Common Multiple of two or more numbers is a number which is a multiple of each of them; thus, 24 is & common multiple of 2, 3, and 4.

126. The Least Common Multiple of two or more numbers is the least number which is a multiple of each of them; thus, 12 is the least common multiple of 2, 3, and 4.

NOTE.-The least common multiple may be represented by the initials I C. M.

PRINCIPLES. 1. A multiple of a number is exactly divisible by that nuniber.

2. A multiple of a number must contain all the prime factors of that number.

3. A common multiple of two or more numbers must com tain all the prime factors of each of those numbers.

4. The ieast common multiple of two or more numbers must contain all the prime factors of each number, and no other factors.

CASE I.

127. When the numbers are small and easily face cored.

FIRST METHOD.

128. This method consists in resolving tbe numbers into their prime factors, and taking the product of all the differ. ent factors."

1. Find the least common multiple of 12, 30, and 70. SOLUTION.We first resolve the numbers OPERATION. into their prime factors. A multiple of 12

12=2X2X3 must contain the factors of 12, 2, 2, 3; a

30—2X3X5 multiple of 30 must contain the factors of 30,

70=2X5X7 2, 3, 6; a multiple of 70 must contain the

2X2X3X6X7=420 factors of 70, 2, 5, 7; hence the common multiple of 12, 30, and 70 must contain all these different factors and no others; therefore 2x2x5x3x7, or 420, is the L. C. M. of 12, 30, and 70 (Prin. 4).

Rule.-I. Resolve the numbers into their prime factors. II. Take the product of all the different factors, using each factor the greatest number of times it occurs in either number.

NOTE. -Any numbers which are divisors of the others may be omitted, since the multiple of the other numbers will be a multiple of these.

Find the least common multiple of 2. 24, 30, and 36.

Ans. 360. 3. 16, 24, and 56.

Ans. 336. 4. 28, 36, and 60.

Ans. 1260. 5. 36, 48, and 84.

Ans. 1008. 6. 63, 72, and 108.

Ans. 1512. 7. 15, 30, 42, and 72.

Ans. 2520. 8. 22, 55, 77, and 110.

Ans. 770. 9. 33, 99, 36, 108, and 135.

Ans. 5940. 10. A has $14, B $15, C $36, and D as many as the least common multiple of the amounts of the others; how many has D?

Ans. $1260.

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OPERATION.

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SECOND METHOD. 129. This method consists in taking out the prime fac tors of the last common multiple and finding their product.

1. Find the least common multiple of 12, 30, and 70. SOLUTION.—Placing the numbers one beside another and dividing by 2, we see that 12:12 50 - 70 2 is a factor of each of them, it is therefore a factor of the L. C. M. (Prin. 3); dividing the

31 6 -15 35 quotients that will contain it by 3, we see 5 2 5 35 that s is a factor of some of the numbers, it is 2 1 7 therefore a factor of the L. C. M. Dividing the next quotients by 5, we see that 5 is a fan

2X3X5X2X7=420 tor of some of them, hence 5 is a factor of the L. C. M.; and the quotients having no other common factor, we see that the factors of the given numbers are 2, 3, 5, 2, and 7, hence their product, which is 420, is the L. C. M. Hence the following

Rule.-I. Write the numbers one beside another, divide by any prime number that will cactly divide two or more, and write the quotients and undivided numbers beneath.

II. Divide the quotients in the same manner, and thus continue until no two numbers in the lowest line have a common factor.

III. Take the product of the divisors and final quotients; the result will be the least common multiple required.

Find the least common multiple of 2. 16, 20, and 30.

Ans. 240. 3. 28, 56, and 84.

Ans. 168. 4. 48, 60, and 30.

Ans. 240. 5. 150, 200, and 250.

Ans. 3000. 6. 40, 96, 100, and 120.

Ans. 2400. V 7. 120, 180, 200, and 240

Ans. 3600. 8. 140, 280, 160, and 320.

Ans. 2240.

CASE II. 130. When the numers are large and cannot be eadily factored.

1. Find the least common multiple of 28 and 63.

SOLUTION.-The greatest common divisor uf these numbers is 7; 28 equals 4 times 7, 28=4X7; 63=9X7 and 63 equals 9 times 7; hence the L. C. M., as found in the first method, is 4 x 7 x 9,

L C. M.=4X7X9

63 which equals 28 multiplied by 63 divided by 7: or the first number multiplied by the second divided by their greatest common aruism.

OPERATION.

=28 XF

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