EXACT DIVISION. 169. The following tests of exact division should be carefully studied and fixed in the memory for future use. PROP. I.-A divisor of any number is a divisor of any number of times that number. Thus, 123 fours. Hence, 12 x 6 3 fours × 6=18 fours. But 18 fours are divisible by 4. Hence, 12 x 6, or 72, is divisible by 4. PROP. II.—A divisor of each of two or more numbers is a divisor of their sum. Thus, 5 is a divisor of 10 and 30; that is, 10: 306 fives. Hence, 10 + 30 But 8 fives are divisible by 5. sum of 10 and 30. = 2 fives and = 8 fives. 2 fives+6 fives Hence, 5 is a divisor of the PROP. III.—A divisor of each of two numbers is a divisor of their difference. Thus, 3 is a divisor of 27 and 15; that is, 27 = 9 threes and 15 5 threes. Hence, 27 159 threes 5 threes = 4 threes. But 4 threes are divisible by 3. divisor of the difference between 27 and 15. Hence 3 is a PROP. IV.—Any number ending with a cipher is divisible by the divisors of 10, viz., 2 and 5. Thus, 370 37 times 10. Hence is divisible by 2 and 5, the divisors of 10, according to Prop. I. PROP. V. Any number is divisible by either of the divisors of 10, when its right-hand figure is divisible by the same. Thus, 498 490 + 8. Each of these parts is divisible by 2. Hence the number 498 is divisible by 2, according to Prop. II. In the same way it may be shown that 495 is divisible by 5. PROP. VI.-Any number ending with two ciphers is divisible by the divisors of 100, viz., 2, 4, 5, 10, 20, 25, and 50. Thus, 8900 = 89 times 100. Hence is divisible by any of the divisors of 100, according to Prop. I. PROP. VII.-Any number is divisible by any one of the divisors of 100, when the number expressed by its two righthand figures is divisible by the same. Thus, 49754900+75. Any divisor of 100 is a divisor of 4900 (Prop. VI). Hence, any divisor of 100 which will divide 75 is a divisor of 4975 (Prop. II). PROP. VIII.—Any number ending with three ciphers is divisible by the divisors of 1000, viz., 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, and 500. Thus, 83000 83 times 1000. Hence is divisible by any of the divisors of 1000, according to Prop. I. PROP. IX.-Any number is divisible by any one of the divisors of 1000, when the number expressed by its three righthand figures is divisible by the same. Thus, 9262592000+ 625. divisor of 92000 (Prop. VIII). Any divisor of 1000 is a Hence, any divisor of 1000 which will divide 625 is a divisor of 92625 (Prop. II). PROP. X.-Any number is divisible by 9, if the sum of its digits is divisible by 9. This proposition may be shown thus: (1.) 486400+80+ 6. = (2.) 100 99 +1= 11 nines + 1. Hence, 400 = 44 nines + 4, and is divisible by 9 with a remainder 4. (3.) 10 = 9 + 1 = 1 nine + 1. Hence, 808 nines + 8, and is divisible by 9 with a remainder 8. (4.) From the foregoing it follows that 400 + 80 + 6, or 486, is divisible by 9 with a remainder 4 + 8 +6, the sum of the digits. Hence, if the sum of the digits is divisible by 9, the number 486 is divisible by 9 (Prop. II). PROP. XI.-Any number is divisible by 3, if the sum of its digits is divisible by 3. This proposition is shown in the same manner as Prop. X; as 3 divides 10, 100, 1000, etc., with a remainder 1 in each case. PROP. XII.—Any number is divisible by 11, if the difference of the sums of the digits in the odd and even places is zero or is divisible by 11. This may be shown thus: (1.) 4928 = 4000 + 900 + 20 + 8. (2.) 1000 91 elevens-1. 2. Hence, 4000=364 elevens-4. (3.) 100 = 9 elevens + 1. Hence, 90081 elevens + 9. (4.) 10 = 1 eleven 1. Hence, 20 = 2 elevens (5.) From the foregoing it follows that 4928 = 364 elevens +81 elevens + 2 elevens-4+9-2+8. But 492+8=11. Hence, 4928364 elevens +81 elevens + 2 elevens + 1 eleven 448 elevens, and is therefore divisible by 11. The same course of reasoning applies where the difference is minus or zero. Hence, etc. EXAMPLES FOR PRACTICE. 170. Find exact divisors of each of the following numbers by applying the foregoing tests: PRIME NUMBERS. PREPARATORY PROPOSITIONS. 171. PROP. I.—All even numbers are divisible by 2 and consequently all even numbers, except 2, are composite. Hence, in finding the prime numbers, we cancel as composite all even numbers except 2. Thus, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, and so on. PROP. II.—Each number in the series of odd numbers is 2 greater than the number immediately preceding it. are Thus, the numbers left after cancelling the even numbers PROP. III-In the series of odd numbers, every THIRD number from 3 is divisible by 3, every FIFTH number from 5 is divisible by 5, and so on with each number in the series. This proposition may be shown thus: According to Prop. II, the series of odd numbers increase by 2's. Hence the third number from 3 is found by adding 2 three times, thus: From this it will be seen that 9, the third number from 3, is composed of 3, plus 3 twos, and is divisible by 3 (Prop. II); and so with the third number from 9, and so on. By the same course of reasoning, each fifth number in the series, counting from 5, may be shown to be divisible by 5; and so with any other number in the series; hence the following method of finding the prime numbers. 172. PROB.-To find all the Prime Numbers from 1 to any given number. Find all the prime numbers from 1 to 63. EXPLANATION.-1. Arrange the series of odd numbers in lines, at convenient distances from each other, as shown in illustration. 2. Write 3 under every third number from 3, 5 under every fifth number from 5, 7 under every seventh number from 7, and so on with each of the other numbers. 3. The terms under which the numbers are written are composite, and the numbers written under are their factors, according to Prop. III. All the remaining numbers are prime. Hence all the prime numbers from 1 to 63 are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61. EXAMPLES FOR PRACTICE. 173. 1. Find all the prime numbers from 1 to 95. 6. Show by an example that every seventh number from seven, in the series of odd numbers, is divisible by seven. |