math336_hw2_solns - n = p e divides p ( p 2 ) ... ( p e-1...

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Homework 2 Partial Solutions Question 5A: # 7 n is composite, n > 4. We want to show n | ( n - 1)! Let p be a prime dividing n. Let e be the highest power of p diving n . If p e < n , then clearly p e | ( n - 1)!. Also, n p e 6 = p e and n p e < n , so p e n p e | ( n - 1)! or n | ( n - 1)!. If p e = n , then e > 1 since n is not prime, and thus 2 p < n since n > 4. But then
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Unformatted text preview: n = p e divides p ( p 2 ) ... ( p e-1 )(2 p ), which in turn divides ( n-1)!. Question 5B: # 8 6 -5 (mod 11), so 6 e (-5) e =-5 e (mod 11) since e is odd. Then 5 e + 6 e 5 e-5 e = 0 (mod 11). 1...
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This note was uploaded on 06/21/2008 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell University (Engineering School).

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