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Spring 2007 - Eggers' Class - Exam 2

# Spring 2007 - Eggers' Class - Exam 2 - Math 109 Midterm...

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Unformatted text preview: Math 109 Midterm Exam 2 Solution 1. (4 points) Using the Euclidean algorithm, find gcd(3589 , 2627) and find integers m and n such that gcd(3589 , 2627) = 3589 m + 2627 n . 3589 = 2627 + 962 37 = 11(2627)- 30(3589- 2627) = 41(2627)- 30(3589) 2627 = 2(962) + 703 37 = 11[2627- 2(962)]- 8(962) = 11(2627)- 30(962) 962 = 703 + 259 37 = 3(703)- 8(962- 703) = 11(703)- 8(962) 703 = 2(259) + 185 37 = 3[703- 2(259)]- 2(259) = 3(703)- 8(259) 259 = 185 + 74 37 = 185- 2(259- 185) = 3(185)- 2(259) 185 = 2(74) + 37 37 = 185- 2(74) 74 = 22(37) Thus, gcd(3589 , 2627) = 37 and 41(2627)- 30(3589) = gcd(3589 , 2627). 2. (6 points) Let n be a positive integer. Prove that 7 divides 6 n + 1 if and only if n is odd. [Hint: 6 ≡ - 1 mod 7.] Note that 7 divides 6 n + 1 if and only if 6 n ≡ - 1 mod 7 and observe that since 6 ≡ - 1 mod 7, it follows that 6 n ≡ (- 1) n mod 7. Thus, it suffices to prove that (- 1) n ≡ - 1 mod 7 if and only if n is odd. Since (- 1) n = 1 if and only if n is odd, the result follows immediately.is odd, the result follows immediately....
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Spring 2007 - Eggers' Class - Exam 2 - Math 109 Midterm...

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