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Unformatted text preview: Math 135, Winter 2010 Solutions to Assignment 5 Due 8:30am on February 24th in the drop box outside the Tutorial Centre (MC 4067) [3 pts] 1) Find the inverse, or prove that one doesn’t exist, of (a) [10] in Z 50 , and Solution: We need to find [ x ] such that [10][ x ] = [1]. This is the same as solving the linear congruence 10 x ≡ 1 (mod 50). But by the Linear Congru ence Theorem, the congruence has a solution if and only if gcd(10 , 50)  1. Since gcd(10 , 50) = 10 1, there is no solution to the congruence. So [10] does not have an inverse in Z 50 . (b) [11] in Z 50 , Solution: Similarly to the previous problem, we need to solve the congruence 11 x ≡ 1 (mod 50). To do this, we solve the Diophantine equation 11 x +50 y = 1 using the Extended Euclidean Algorithm: x y r 1 11 1 50 1 11 4 1 6 5 1 5 9 2 1 So a particular solution is x = 9, y = 2. This yields a general solution of x ≡  9 (mod 50), y ≡ 2 (mod 11). Thus, the solution of the linear congruence is x ≡  9 (mod 50), or x ≡ 41 (mod 50), and we have [11] 1 = [41]. [6 points ] 2) Use the Chinese Remainder Theorem to solve the congruence x 2 ≡ x + 2 (mod 77). (Do NOT solve by trying all 77 congruence classes.) Solution: By Proposition 3.64, we know that x 2 ≡ x + 2 (mod 77) if and only if x 2 ≡ x + 2 (mod 7) and x 2 ≡ x + 2 (mod 11). We construct tables 1 as follows to solve these congruences: Modulo 7 x ≡ 1 2 3 4 5 6 x 2 ≡ 1 4 2 2 4 1 x + 2 ≡ 2 3 4 5 6 1 Modulo 11 x ≡ 1 2 3 4 5 6 7 8 9 10 x 2 ≡ 1 4 9 5 3 3 5 9 4 1 x + 2 ≡ 2 3 4 5 6 7 8 9 10 1 So x 2 ≡ x + 2 (mod 7) if and only if x ≡ 2 (mod 7) or x ≡ 6 (mod 7), and x 2 ≡ x +2 (mod 11) if and only if...
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This note was uploaded on 04/13/2010 for the course MATH 135 taught by Professor Andrewchilds during the Spring '08 term at Waterloo.
 Spring '08
 ANDREWCHILDS
 Math

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