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# sol4 - Math 465 Solution Set 4 1 Solve the simultaneous...

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Math 465 Solution Set 4 Ae Ja Yee 1. Solve the simultaneous congruences x 3 (mod 6) , x 5 (mod 35) , x 7 (mod 143) . Solution. Since 1 = 6 · 6 + 35 · ( - 1), we obtain x 5 · 6 · 6 + 35 · ( - 1) · 3 75 (mod 210) . We now solve x 75 (mod 210) , x 7 (mod 143) . We apply Euclidean algorithm to write (210 , 143) = 1 as a linear combination of 210 and 143. 210 = 143 · 1 + 67 143 = 67 · 2 + 9 67 = 9 · 7 + 4 9 = 4 · 2 + 1 so x 210 · ( - 32) · 7 + 143 · 47 · 75 457035 6585 (mod 30030) . 2. Prove that x b 1 (mod m 1 ) x b 2 (mod m 2 ) is solvable if and only if ( m 1 , m 2 ) | b 1 - b 2 . In this case, prove that the solution is unique modulo [ m 1 , m 2 ]. Solution. Let d = ( m 1 , m 2 ). From the first congruence, x = m 1 k + b 1 for any integer k . Plug x into the second congruence. Then m 1 k + b 1 b 2 (mod m 2 ) , m 1 k b 2 - b 1 (mod m 2 ) . Thus the original system of congruences is solvable if and only if the second congruence is solvable, which is equivalent to d | b 2 - b 1 . 1

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3. Using Euler’s Theorem, find the least nonnegative residue modulo m of each integer n below.
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