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mid-solutions

mid-solutions - Midterm Mathematics 312 Section 101...

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Midterm. Mathematics 312, Section 101 Thursday, October 21, 2010. Instructor: Z. Reichstein. Problem 1: (6 marks) Use mathematical induction to show that 2 n > n 2 for every integer n 5. Solution: Base case. For n = 5, 2 n = 32 and n 2 = 25, so the desired inequality holds. For the induction step assume (1) 2 n > n 2 for some n 5. We need to show that 2 n +1 > ( n + 1) 2 . Multiplying both sides of (1) by 2, we obtain 2 n +1 > 2 n 2 . It thus remains to show that 2 n 2 ( n + 1) 2 for every n 5. Indeed, 2 n 2 > 36 25 n 2 = (1 + 1 5 ) 2 n 2 (1 + 1 n ) 2 n 2 = ( n + 1) 2 , as claimed. This completes the proof. Problem 2: (6 marks) Find all integer solutions of the linear equation ax + (3 a + 1) y = 1 . Here a is a given positive integer and x , y are unknowns. Solution: One solution is clearly x 0 = - 3, y 0 = 1. Since it exists, gcd( a, 3 a + 1) divides 1 and hence, equals 1. The general solution is thus x = - 3 + (3 a + 1) n , y = 1 - an , as n ranges over the integers. Remark: The fact that gcd(3 a + 1 , a ) = 1 can also be checked directly as follows: gcd(3 a + 1 , a ) = gcd(3 a + 1 - 3 a, a ) = gcd(1 , a ) = 1. The above slution does not rely

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