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Unformatted text preview: MATH 135 Algebra, Solutions to Assignment 6 1: Find the prime factorization of each of the following integers. (a) 30! Solution: First, let us describe a method for finding the exponent of a prime p in the prime factorization of n ! for any positive integer n . Note that n ! is the product of the numbers 1 , 2 , 3 , ,n . The multiples of p that occur in this list are p, 2 p, 3 p, , j n p k , so there are j n p k multiples of p in the list. Similarly, there are j n p 2 k multiples of p 2 in the list and j n p 3 k multiples of p 3 and so on. Thus the exponent of the prime p in the prime factorization of n ! is equal to j n p k + j n p 2 k + j n p 3 k + . Using the above rule, the exponent of 2 in 30! is 30 2 + 30 4 + 30 8 + 30 16 = 15 + 7 + 3 + 1 = 26, the exponent of 3 is 30 3 + 30 9 + 30 27 = 10 + 3 + 1 = 14, the exponent of 5 is 30 5 + 30 25 = 6 + 1 = 7, the exponent of 7 is 30 7 = 4, the exponent of 11 is 30 11 = 2, the exponent of 13 is 30 13 = 3, and the exponents of 17, 19, 23 and 29 are all equal to 1. Thus we have 30! = 2 26 3 14 5 7 7 4 11 2 13 2 17 19 23 29 . (b) 30 10 Solution: Using our rule from part (a) we find that 10! = 2 5+2+1 3 3+1 5 2 7 , and 20! = 2 10+5+2+1 3 6+2 5 4 7 2 11 13 17 19 and so 30 10 = 30! 10!20! = 2 26 3 14 5 7 7 4 11 2 13 2 17 19 23 29 (2 8 3 4 5 2 7)(2 18 3 8 5 4 7 2 11 13 17 19) = 3 2 5 7 11 13 23 29 . (c) 2 36 1 Solution: Recall that a 2 b 2 = ( a b )( a + b ), a 3 b 3 = ( a b )( a 2 + ab + b 2 ) and a 3 + b 3 = ( a + b )( a 2 ab + b 2 ). Use these rules repeatedly to get 2 36 1 = (2 18 1)(2 18 + 1) = (2 9 1)(2 9 + 1)(2 6 + 1)(2 12 2 6 + 1) = (2 3 1)(2 6 + 2 3 + 1)(2 3 + 1)(2 6 2 3 + 1)(2 2 + 1)(2 4 2 2 + 1)(2 12 2 6 + 1) = 7 73 9 57 5 13 4033 = 7 73 3 2 3 19 5 13 4033 . Note that 73 is prime, since b 73 c = 8, and none of the primes 2, 3, 5, 7 is a factor of 73. To determine whether 4033 is prime, we test every prime p with p b 4033 c = 63 to see if it is a factor. Using the Sieve of Eratosthenes, we find that the primes we need to check are 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , and when we test these we find that 37 is a factor, indeed 4033 = 37 109. Note that 109 is prime, since b 109 c = 10 and none of the primes 2, 3, 5, 7 is a factor of 109. Thus we obtain the prime factorization 2 36 1 = 3...
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This note was uploaded on 05/04/2010 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.
 Fall '08
 ANDREWCHILDS
 Algebra, Integers

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