midterm2_sol - Faculty of Mathematics University of...

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Faculty of Mathematics University of Waterloo MATH 135 MIDTERM EXAM #2 Fall 2005 Monday 14 November 2004 19:00 – 20:15 Solutions 1. Solve the system of congruences [8] x 40 (mod 131) x 7 (mod 18) Solution: Since x 40 (mod 131), then x = 40 + 131 y for some y Z . Substituting into the second congruence, we obtain 40 + 131 y 7 (mod 18) 131 y ≡ - 33 (mod 18) 5 y ≡ - 33 (mod 18) since 131 5 (mod 18). We can solve this linear congruence in a number of ways, for instance by inspection or by converting to the linear Diophantine equation 5 y + 18 z = - 33. Another approach would be to notice that - 33 ≡ - 15 (mod 18), so 5 y ≡ - 15 (mod 18). Since gcd(5 , 18) = 1, then we can divide out the common factor of 5 to obtain y ≡ - 3 (mod 18). Thus, y = - 3 + 18 z for some z Z . Therefore, x = 40 + 131( - 3 + 18 z ) = - 353 + 2358 z so x ≡ - 353 (mod 2358) or x 2005 (mod 2358). 2. Prove that, for all positive integers n , [7] 1 1 · 3 + 1 3 · 5 + ··· + 1 (2 n - 1)(2 n + 1) = n 2 n + 1 Solution: We prove this result by induction n . Base Case If n = 1, the left side equals 1 (1)(3) = 1 3 and the right side equals 1 2(1) + 1 = 1 3 , so the result is true for n = 1. Induction Hypothesis
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midterm2_sol - Faculty of Mathematics University of...

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