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Midterm 2 Solution

Midterm 2 Solution - 2 Problem 3 Use mathematical induction...

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Math 243, Spring 2006, Professor Callahan Test #2, Thu–Fri, Apr. 13–14. Note 1: This test is closed book. You may use one 8 1 2 × 11 sheet of notes (both sides). Note 2: Show your work. Clarity counts. If I can’t follow your reasoning I can’t give credit. Problem 1: Evaluate 7 1172 mod 31. Answer: We know that 31 is prime, so by Fermat’s Little Theorem 7 30 1 (mod 31). Knowing that 1172 = 30 · 39 + 2, we get 7 1172 = 7 30 · 39+2 = 7 30 · 39 · 7 2 = (7 30 ) 39 · 7 2 1 39 · 7 2 49 18 (mod 31) . 1
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Problem 2: Find the value of x between 0 and 280 that satisfies x 2 (mod 5) x 2 (mod 7) x 3 (mod 8) . Answer: We use the Chinese Remainder Theorem. With m 1 = 5, m 2 = 7 and m 3 = 8 we have M 1 = m 2 m 3 = 56 , M 2 = m 1 m 3 = 40 , M 3 = m 1 m 2 = 35 . Then M 1 mod m 1 = 56 mod 5 = 1 , so the inverse is y 1 = 1. M 2 mod m 2 = 40 mod 7 = 5 . We note that 3 · 5 = 15 1 (mod 7), so the inverse is y 2 = 3. M 3 mod m 3 = 35 mod 8 = 3 . We note that 3 · 3 = 9 1 (mod 8), so the inverse is y 3 = 3. Then we take a 1 M 1 y 1 + a 2 M 2 y 2 + a 3 M 3 y 3 = 2 · 56 · 1 + 2 · 40 · 3 + 3 · 35 · 3 = 667 107 (mod 280) . Thus x = 107. 2
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Problem 3: Use mathematical induction to show that 5 divides
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Unformatted text preview: 2 Problem 3: Use mathematical induction to show that 5 divides n 5-n whenever n is a nonnegative integer. Answer: First we check the smallest case, with n = 0. Then n 5-n = 0, which is divisible by 5. Now we assume that n 5-n is divisible by 5. Then ( n + 1) 5-( n + 1) = n 5 + 5 n 4 + 10 n 3 + 10 n 2 + 5 n + 1-n-1 = ( n 5-n ) + 5 n 4 + 10 n 3 + 10 n 2 + 5 n = ( n 5-n ) + 5( n 4 + 2 n 3 + 2 n 2 + n ) . The frst part is a multiple o± 5 by our inductive hypothesis, and the second part is obviously a multiple o± 5, so the whole thing is a multiple o± 5. 3 Problem 4: Use the geometric sum formula to evaluate 2 + 4 3 + 8 9 + ··· + 2 p 2 3 P 8 . Answer: This is 2 + 2 · p 2 3 P + 2 · p 2 3 P 2 + ··· + 2 p 2 3 P 8 = 2(2 / 3) 9-2 (2 / 3)-1 = 38342 6561 ≈ 5 . 84393 . 4...
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