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189. Every number is divisible by 11, in which the sum of the digits in the odd places is equal to the sum of the digits in the even places, or in which the difference of their sums can be exactly divided by 11. Thus, 8305, in which 3+5=8+0, and 628001, in which 2 +0+1 and 6 + 8 +0 differ by 11, are each divisible by 11.

190. Every number divisible by two or more numbers, which are prime to each other, is divisible by their product. For, being prime to each other, dividing by one of the numbers does not cancel the others as factors. Thus, 770, being divisible by 2, 5, and 7, which are prime to each other, is divisible by 70, their product.

191. Every even number, the sum of whose digits 6 will exactly divide, is divisible by 6. For being even, it is divisible by 2 (Art. 183), and its digits being divisible by twice 3, or 6, are evidently divisible by once 3, so that the number is also divisible by 3; and as the 2 and the 3 are prime to each other, the number is divisible by their product, or 6 (Art. 190). Thus, 174, 6312, are each divisible by 6.

192. Every number terminating with 0 or 5 that 3 will exactly divide, is divisible by 15, and every number that 9 will exactly divide, is divisible by 45. For, terminating with 0 or 5, it is divisible by 5 (Art. 184), and, as 3 and 9 are each prime to 5, if it can be exactly divided by 3 or by 9, it must be divisible by 5 × 3: 15, or by 5 × 9 45. Thus, 75, which 3 will exactly divide, is divisible by 15; and 90, which 9 will exactly divide, is divisible by 45.

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DIVISORS OR MEASURES.

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193. A divisor or measure of a number is any number that will divide it without a remainder. Thus, 3 is a divisor or measure of 6, and 5 a divisor or measure of 10.

194. To find all the divisors or measures of a number.

Ex. 1. Required all the divisors of 60.

Ans. 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60

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Ans. 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.

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, any num

ber is divisible by 1, and every composite number by its Now 1,

prime factors, and every product these factors can form. being a divisor of every number, is a divisor of 60, and, since 2 enters twice as a factor into 60, it is evident that 2, and 4 = 2 × 2, are also divisors of 60. These divisors we arrange on a horizontal line, and determine the other divisors by multiplying those on this line by the factor 3, for the second line of divisors, by 5 for the third line, and by 3 x 5 for the fourth line, and thus obtain all the possible divisors of the given number.

The whole number of divisors is 12, which corresponds to the product arising from multiplying together the exponents, each increased by 1, of the different prime factors of 60; thus, of the different prime factors, since 2 enters twice, its exponent is 2, +1: 3; enters once, its exponent is 1, +1 = 2; 5 enters once, its exponent is 1, +1 2; and 3 x 2 x 2 = = 12, the number of divisors. The same holds true in all cases. Hence, in any composite number,

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TO FIND THE NUMBER OF DIVISORS. ·Multiply together the exponents, each increased by 1, of the different prime factors of the given number, and the product will be the number of divisors required. And

TO FIND THE SEVERAL DIVISORS.-Form from the prime factors of the number all the products possible, and these factors (including 1) and products will be the divisors required.

EXAMPLES.

2. What are the divisors of 72?

Ans. 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72.

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195. A common divisor or measure of two or more numbers is any number that will divide them without a remainder; thus, 2 is a common divisor of 4, 8, and 10.

196. A common divisor of two numbers is a divisor of their

sum, and also of their difference. of 12 and 18, is a divisor of their ference, 6.

Thus, 6, a common divisor sum, 30, and of their dif

197. A common divisor of the remainder and the divisor is a divisor of the dividend. Thus, in a division having 8 for a remainder, 16 for divisor, and 24 for dividend, 8, a common divisor of 8 and 16, is also a divisor of the 24.

198. To find all the divisors common to two or more numbers.

Ex. 1. Required all the common divisors of 45 and 135. Ans. 1, 3, 5, 9, 15, and 45.

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Resolving the giver numbers into their prime factors, we find they have of these 3, 3, and 5 in common, and these COMMON

3 × 3 × 5. prime factors with 1, and all the products we are able to form from them (Art. 194), give

Ans. 1, 3, 5, 9, 15, and 45.

all the common divisors required. When only the number of common divisors is required, it may readily be found by multiplying together the exponents, each increased by 1, of the different COMMON prime factors. (Art. 194.)

EXAMPLES.

2. What are the common divisors of 51, 153, and 255 ? 3. Required the several common divisors of 180 and 360. Ans. 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, and 180.

4. How many common divisors have 2025, 6075, and 8100? Ans. 15.

5. How many common divisors have 4500 and 9000?

Ans. 36.

THE GREATEST COMMON DIVISOR OR MEASURE.

199.

The greatest common divisor or measure of two or more numbers is the greatest number that will divide each of

them without a remainder. Thus, 4 is the greatest common divisor of 8, 12, and 16.

200.

numbers.

Ex. 1.

To find the greatest common divisor of two or more

Required the greatest common divisor or measure

of 24 and 88.

FIRST OPERATION.

24

= 2 X 2 X 2 X 3

= 8. Ans.

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Ans. 8.

Resolving the numbers into their prime factors, thus, 24 2 X 2 X 2 88 X 3, and 88 2 X 2 X 2 X 11. 2 X 2 X 2 X 11 = 88, we find the factors 2 x 2 x 2 2 X 2 X 2 are common to both. Since only these common factors, or the product of two or more of such factors, will exactly divide both numbers, it follows that the product of all their common prime factors must be the greatest factor that will exactly divide both of them. Therefore, 2 × 2 X 2 = 8, the greatest common divisor required.

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The same result may be obtained by a sort of trial process, as by the second operation.

SECOND OPERATION.

24) 88(3
72

16) 24 (1
16
8) 16 (2
16

It is evident, since 24 cannot be exactly divided by a number greater than itself, if it will also exactly divide 88, it will be the greatest common divisor sought. But, on trial, we find 24 will not exactly divide 88, there being a remainder, 16. Therefore 24 is not a common divisor of the two numbers.

We know that a common divisor of 16 and 24 will, also, be a common divisor of 88 (Art. 197). We next try to find that divisor. It cannot be greater than 16. But 16 will not exactly divide 24, there being a remainder, 8; therefore 16 is not the greatest common divisor.

As before, the common divisor of 8 and 16 will be the common divisor of 24 and 88 (Art. 197); we make trial to find that divisor, knowing that it cannot be greater than 8, and find 8 will exactly divide 16. Therefore 8 is the greatest common divisor required.

THIRD OPERATION.

24) 88(3

72

16

The last method may be often contracted, if there should be observed to be any prime factor in a remainder which is not common to the preceding divisor, by canceling said factor. Thus, in the third operation, the factor 2 being found in the remainder 16 once more than in the di

8) 24 (3 visor 24, we cancel one 2 from 16, and, having left the composite factor 8, we divide 24 by that factor. There being no remainder, 8 is the greatest common divisor, as before obtained.

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RULE 1. Resolve the given numbers into their prime factors. The product of all the factors common to the several numbers will be the greatest common divisor. Or,

RULE 2. - Divide the greater number by the less, and if there be a remainder divide the preceding divisor by it, and so continue dividing until nothing remains. The last divisor will be the greatest common di

visor.

NOTE. When the greatest common divisor is required of more than two numbers, find it of two of them, and then of that common divisor and of one of the other numbers, and so on for all the given numbers. The last common divisor will be the greatest common divisor required.

Another method is to divide the numbers by any factor common to them all; and so continue to divide till there are no longer any common factors; and the product of all the common factors will be the greatest common divisor required.

EXAMPLES.

2. What is the greatest common divisor of 56 and 168? 3. What is the greatest common divisor of 96 and 128? 4. What is the greatest common measure of 57 and 285? Ans. 57.

5. What is the greatest common measure of 169 and 175? 6. What is the greatest common measure of 175 and 455? Ans. 35. 7. What is the greatest common divisor of 169 and 866? Ans. 1. 8. What is the greatest common measure of 47 and 478? Ans. 1. 9. What is the greatest common measure of 84 and 1068? Ans. 12. 10. What is the greatest common divisor of 75 and 165? Ans. 15. 11. What is the greatest common measure of 78, 234, and 468? Ans. 78.

12. I have three fields; one containing 16 acres; the second, 20 acres; and the third, 24 acres. Required the largest-sized lots, containing each an exact number of acres, into which the whole can be divided. Ans. 4 acre lots.

13. A farmer has 12 bushels of oats, 18 bushels of rye, 24 bushels of corn, and 30 bushels of wheat. Required the largest bins, of uniform size, and containing an exact number of bushels, into which the whole can be put, each kind by itself, and all the bins be full.

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