135_8_-_Solutions

135_8_-_Solutions - Assignment 8 Solutions 1 Solve the...

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Assignment 8 – Solutions 1. Solve the following pairs of congruences. 15 x 4 (mod 26) 24 x 6 (mod 63) Solution: First we solve the linear congruence 15 x = 4 (mod 26). By inspection, one solution is x = 2 and we have gcd(15 , 26) = 1, and so by the Linear Congruence Theorem, the general solution is x 2 (mod 26). Next we solve 24 x 6 (mod 63). We need 24 x + 63 y = 6. The Euclidean Algorithm gives 63 = 2 · 24 + 15 , 24 = 1 · 15 + 9 , 15 = 1 · 9 + 6 , 9 = 1 · 6 + 3 , 6 = 2 · 3 + 0 so we have gcd(24 , 63) = 3, and then Back-Substitution gives 1 , - 1 , 2 , - 3 , 8 so we have (24)(8) + (63)( - 3) = 3. Multiply both sides by 2 to get (24)(16) + (63)( - 6) = 6. Thus one solution is x = 16. Note that 63 3 = 21, so by the Linear Congruence Theorem, the general solution to the congruence 24 x 6 (mod 63) is x 16 (mod 21). Thus the original pair of congruences is equivalent to the pair of congruences x 2 (mod 26) x 16 (mod 21) For x to be a solution we need x = 2 + 26 r and x = 16 + 21 s for some integers r and s , so we must have 2 + 26 r = 16 + 21 s , that is 26 r - 21 s = 14. The Euclidean Algorithm gives 26 = 1 · 21 + 5, 21 = 4 · 5 + 1 so we have gcd(26 , 21) = 1, and then Back-Substitution gives 1, - 4, 5, so we have (26)( - 4) - (21)( - 5) = 1. Multiply both sides by 14 to get (26)( - 56) - (21)( - 70) = 14. Thus one solution is ( r, s ) = ( - 56 , - 70) and the general solution is ( r, s ) = ( - 56 , - 70) + k (21 , 26), k Z , so we have r ≡ - 56 7 (mod 21). Thus one solution to the pair of congruences is x = 2 + 26 r = 2 + 26 · 7 = 184, and by the Chinese Remainder Theorem, the general solution is x

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