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Unformatted text preview: 1 MATH 135 F 2011: Assignment 5 Solutions 1. The Chinese Remainder Theorem deals with the case where the moduli are coprime. We now inves tigate what happens if the moduli are not coprime. (a) Consider the following two systems of linear congruences: A : n ≡ 2 (mod 12) n ≡ 10 (mod 18) B : n ≡ 5 (mod 12) n ≡ 11 (mod 18) Determine which one has solutions and which one has no solutions. For the one with solutions, give the complete solutions to the system. For the one with no solutions, explain why no solutions exist. Solution. For system A , the first congruence implies n = 2 + 12 x for some x ∈ Z . Substitute this into the second congruence to get 2 + 12 x ≡ 10 (mod 18), or 12 x ≡ 8 (mod 18). Since gcd(12 , 18) = 6 which does not divide 8, by LCT 1, this does not have a solution. For system B , the first congruence implies n = 5 + 12 x for some x ∈ Z . Substitute this into the second congruence to get 5 + 12 x ≡ 11 (mod 18), or 12 x ≡ 6 (mod 18). Since gcd(12 , 18) = 6 which divides 6, there is a solution. We can use EEA or by inspection to see that x = 2 is one solution, hence by LCT 1, the complete solution to the second congruence is x ≡ 2 (mod 18 / 6) or x ≡ 2 (mod 3). So x = 2 + 3 y for some y ∈ Z . Substitute back into n to get n = 5 + 12(2 + 3 y ) = 29 + 36 y . So the complete solution is n ≡ 29 (mod 36). (b) Let a 1 , a 2 be integers, and let m 1 , m 2 be positive integers. Consider the following system of linear congruences S : n ≡ a 1 (mod m 1 ) n ≡ a 2 (mod m 2 ) Based on your observations in part (a), complete the following two propositions.Based on your observations in part (a), complete the following two propositions....
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 Spring '08
 ANDREWCHILDS
 Math, Algebra, Remainder Theorem, Remainder, Congruence, Prime number, Wolczuk

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