28146 is divisible by 3, for the sum of its digits is 21, a number divisible by 3. 1342 is not divisible by 3, the sum of its digits being 10, a number not divisible by 3. Since 2 and 3 are the factors of 6, A number is divisible by 6 if it is divisible by 2 and by 3. A number is divisible by 12 if it is divisible by 3 and by 4. Take any number, such that the difference between the sums of the alternate digits is some multiple of 11. (Note that O is a multiple of 11. 0× 11=0.) Take, for instance, 8253817. 7+8+5+8=28. 1+3+2=6. 28-6=22 = 2 x 11. 8253817-7+10+800+3,000+50,000+200,000+8,000,000. 50,000 = 599×99+ 500. 500 5×99+ = 7 . 10 8 . 30 5 . 20 200,000 2,000×99+ 2,000. 2,000 29×99+ 8,000,000 = 80,ØØØ×99+80,000. 80,000 = 800×99+800. 800 = $×99+ 8 Striking out the multiples of 11 (8 × 99, 30 × 99, 500 × 99, 5× 99, etc.), we have left 7+8+5+8+10+30+20; 7, 8, 5, and 8 being one set of alternate digits of 8253817, the tens' digits of 10, 30, and 20 being the other set. We took a number such that the difference between the sums of the alternate digits should be a multiple 11, so 1+3+2+10+30 +20 differs from 7+8+5+8+10+30 +20 by a multiple of 11, and if 1+3+2+10+30+20 be divisible by 11,7+8+5+8 +10+30+20 is also divisible by 11. But 1+3+2+ 10+30 + 2011 +33 +22, a quantity divisible by 11, so 7+8+5+8+10+30 +20 is divisible by 11. The two parts into which we have separated 8253817 are each divisible by 11, therefore 8253817 is divisible by 11. Put in general terms, A number is divisible by 11 if the difference between the sums of its alternate digits is divisible by 11. The sum of one set of alternate digits of 22346 (6+3+2) is 11, the sum of the other set (4+2) is 6. 11-6 5. 5 is not divisible by 11, therefore 22346 is not divisible by 11. The following corollary may be omitted for the present, and taken up later in connection with Problem 271, page 151. COROLLARY. If the sum of the alternate digits beginning with the units be greater than the sum of the alternate digits beginning with the tens, the remainder, after dividing the difference of these two sums by 11, will be the same as the remainder after dividing the number by 11. If the sum of the alternate digits beginning with the units be less than the sum of the alternate digits beginning with the tens, and if the remainder, after dividing the difference between these two sums by 11, be subtracted from 11, this latter remainder will be the same as the remainder after dividing the number by 11. FROM PRACTICE TABLE No. 1. 23. Which of the numbers a-g are divisible by each of the following: 2, 3, 4, 5, 6, 8, 9, 10, 11, 12? 24. Which of the first five numbers a-m are divisible by 3, 9, and 11, respectively? No other two numbers are factors of 12, except 12 itself and 1. 2 × 3 = 6. 2 and 3 are factors of 6. Substituting 2 x 3 for 6 in 2 x 6 = 12, 2 x 2 x 3 = 12. 2, 2, and 3 are factors of 12. 2 x 2 = 4. Substituting 2 × 2 for 4 in 3 x 4 = 12, 3 x 2 x 2 = 12. 3, 2, and 2 are factors of 12; the same result as before. There are no numbers which are factors of 2 or 3, except the numbers themselves and 1. The prime factors of 56 are 2, 2, 2, and 7. 2 and 7 are common factors of 56 and 98. 2 × 7 14. 14 is the greatest common factor of 56 and 98. Find the greatest common factor of 611 and 1363. These numbers cannot be so readily resolved into their factors, and another method may be advantageously employed. 611) 1363 (2 1222 141) 611 (4 First divide the greater number by the less. The number required being a factor of 611 is also a factor of 1222, a multiple of 611; being a factor of 1363 and of 1222, it is a factor of their difference, 141. Divide the smaller of the two given numbers by the remainder, 141. The number required being a factor of 141 is also a factor of 564, a multiple of 141; being a factor of 611 and of 564, it is a factor of their difference, 47. Divide the last divisor, 141, by the last remainder, 47. 47 is a factor of 141. The number required is a common factor of 47 and 141. Therefore, the number required is either 47 or a factor of 47. But 47 being a factor of 141 is a factor of 564, a multiple of 141; being a factor of 564 and of 47, it is a factor of their sum, 611; being a factor of 611 it is a factor of 1222, a multiple of 611; being a factor of 1222 and of 611 it is a factor of their sum, 1363. We have now proved that 47, or a factor of 47, is the greatest common factor of 611 and 1363, and that 47 itself is a common factor of the same two numbers. Hence, 47 is the greatest common factor required. In the above example, after the first division, we have the numbers 141 and 611, and are required to find the greatest common factor of these two numbers. We can readily see that 3 is a factor of 141, but not of 611. Not being a common factor of 141 and 611, we may omit the factor 3 from 141 without affecting the result. Thus, 141 47) 611 (13 47 141 Find the greatest common factor of 1547, 1729, and 1677. 91 is the greatest common factor of 1547 and 1729. 13 is the greatest common factor of 91 and 1677. 13 is the greatest common factor of 1547, 1729, and 1677. FROM PRACTICE TABLE No. 1. 25. What are the prime factors of each of the first thirteen a-b? 26. Of 14-25, a-c? 27. What is the greatest common factor of wx and yz? 28. Of t-w and x-z? 29. Of a-c, d-g, and h-l, in each of the first ten lines? |