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Unformatted text preview: MATH 135, Fall 2010 Solution of Assignment #5 Problem 1 . Find the prime factorization of each of the following integers. (a) 20!. (b) ( 20 5 ) . Solution. (a) First, let us describe a method for finding the exponent of a prime p in the prime factoriza tion 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 20! is 20 2 + 20 4 + 20 8 + 20 16 = 10+5+2+1 = 18, the exponent of 3 is 20 3 + 20 9 = 6+2 = 8, the exponent of 5 is 20 5 = 4, the exponent of 7 is 20 7 = 2, and the exponents of 11, 13, 17, and 19 are all equal to 1. Thus we have 20! = 2 18 ·...
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This note was uploaded on 07/25/2011 for the course MATH 135 taught by Professor Andrewchilds during the Spring '08 term at Waterloo.
 Spring '08
 ANDREWCHILDS
 Algebra, Integers

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