210hw8solns - Math 210 Homework 8 Solutions 1. Solve the...

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Math 210 Homework 8 Solutions 1. Solve the system of congruences: 7 x + 2 y 3 (mod 15) 9 x + 4 y 6 (mod 15) Solution. Multiply the first congruence by 2 to get 14 x + 4 y 6 (mod 15) and then subtract the second congruence to get 5 x 0 (mod 15) This implies 5 x = 15 k , so x = 3 k , so x 0 (mod 3), which means x 0 , 3 , 6 , 9 , 12 (mod 15) . Now plug each of those into one of the original congruences (say the first). x 0 (mod 15) = 0 + 2 y 3 (mod 15) = 2 y 3 (mod 15) x 3 (mod 15) = 21 + 2 y 3 (mod 15) = 2 y ≡ - 18 (mod 15) x 6 (mod 15) = 42 + 2 y 3 (mod 15) = 2 y ≡ - 39 (mod 15) x 9 (mod 15) = 63 + 2 y 3 (mod 15) = 2 y ≡ - 60 (mod 15) x 12 (mod 15) = 84 + 2 y 3 (mod 15) = 2 y ≡ - 81 (mod 15) Each of these can be solved for y : 2 y 3 3 + 15 18 (mod 15) = y 9 (mod 15) 2 y ≡ - 18 ≡ - 18 + 30 12 (mod 15) = y 6 (mod 15) 2 y ≡ - 39 ≡ - 39 + 45
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210hw8solns - Math 210 Homework 8 Solutions 1. Solve the...

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