ans55-mt2 - Math 55. Sample Answers to Second Midterm 1....

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Unformatted text preview: Math 55. Sample Answers to Second Midterm 1. (16 points) Prove that n X k =1 k 7 = ( n 8 ) . We have n X k =1 k 7 n X k =1 n 7 = n n 7 = n 8 , so n k =1 k 7 = O ( n 8 ) . We also have n X k =1 k 7 n X k = d n/ 2 e k 7 n 2 n 2 7 = n 8 2 8 , so n k =1 k 7 = ( n 8 ) . Therefore n k =1 k 7 = ( n 8 ) . 2. (20 points) Use the Euclidean algorithm to find gcd(111 , 33) , and also use it to find integers s and t such that gcd(111 , 33) = 111 s + 33 t . To find the gcd, we have: 111 = 33 3 + 12 ; 33 = 12 2 + 9 ; 12 = 9 1 + 3 ; 9 = 3 3 + 0 . Therefore gcd(111 , 33) = 3 . To get the linear combination, we reverse the above steps, as in the book: 3 = 12- 9 1 = 12- (33- 12 2) = 12 3- 33 = (111- 33 3) 3- 33 = 111 3- 33 10 . Therefore s = 3 and t =- 10 . As an alternative approach, we can do as in class: 111 = 111 1 + 33 0 ; 33 = 111 0 + 33 1 ; 111- 33 3 = 12 = 111 1- 33 3 ; 33- 12 2 = 9 = 111 (- 2) + 33 7 ; 12- 9 = 3 = 111 3- 33 10 . 1 2 This again gives s = 3 and t =- 10 . 3. (16 points) A version of Ackermanns function A : N N N is defined recursively by A ( m,n ) = 2 n if m = 0 or n 1; A ( m- 1 ,A ( m,n- 1)) otherwise....
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ans55-mt2 - Math 55. Sample Answers to Second Midterm 1....

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