135_7_-_Solutions

135_7_-_Solutions - Assignment 7 Solutions 1. Prove,...

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Unformatted text preview: Assignment 7 Solutions 1. Prove, without using a calculator, that 641 divides the Fermat number F (5) = 2 2 5 + 1 . Solution: We have to prove that 2 2 5 +1 0 (mod 641); i.e. 2 32 - 1 (mod 641). We therefore need to compute large powers of 2 modulo 641. We note that 641 = 640+1 = 2 7 5+1, so 2 7 5 - 1 (mod 641) (2 7 5) 4 (- 1) 4 (mod 641) 2 28 625 1 (mod 641) 2 28 (- 16) 1 (mod 641) , since 625 - 16 (mod 641)- 2 32 1 (mod 641) 2 32 - 1 (mod 641) . Therefore, 641 | 2 32 + 1 . 2. Find 27 7 9 (mod 11). Solution1: By Fermats little thereom, since 11 6 | 27, 27 10 1 (mod 11). Consider 7 9 (mod 10). Note that 7 2 49 - 1 (mod 10) We have 7 9 (7 2 ) 4 7 (- 1) 4 7 7 (mod 10) Thus 7 9 = 10 q + 7, for some integer q. The original formula 27 7 9 27 10 q +7 27 10 q 27 7 27 7 (mod 11) Since 27 5 (mod 11), 27 7 9 27 7 5 7 3 (mod 11) . Solution2: Since 27 5 (mod 11) we have 27 7 9 5 7 9 (mod 11). We make a list of powers of 5 modulo 11....
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135_7_-_Solutions - Assignment 7 Solutions 1. Prove,...

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