Midterm2Solutions

Midterm2Solutions - Faculty of Mathematics University of...

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Faculty of Mathematics University of Waterloo MATH 135 MIDTERM EXAM #2 Fall 2008 Monday 10 November 2008 19:00 – 20:15 Solutions 1. In each part of this problem, full marks will be given if the correct answer is written in the box. If your answer is incorrect, your work will be assessed for part marks. (a) Is the statement “The linear Diophantine equation 27 x + 42 y = 17 has integer solutions” [2] TRUE or FALSE? Solution FALSE Reason : Since gcd(27 , 42) = 3 and 3 6 | 17, then there are no solutions. (b) Determine gcd(12! , 2 11 3 4 5 7 ). (You may leave your answer in its prime factorization.) [2] Solution 2 10 3 4 5 2 = 2 073 600 Reason: Since 12! = 12(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1) = 2 10 3 5 5 2 7 1 11 1 , then gcd(12! , 2 11 3 4 5 7 ) = gcd(2 10 3 5 5 2 7 1 11 1 , 2 11 3 4 5 7 ) = 2 10 3 4 5 2 taking the smaller of each pair of corresponding exponents. (c) Determine the remainder when 2 22 + 36 52 is divided by 31. [2] Solution The remainder is 9 Reason: Note first that 2 5 32 1 (mod 31). Also, 36 5 (mod 31) and 5 3 125 1 (mod 31). Thus, 2 22 + 36 52 2 22 + 5 52 (mod 31) (2 5 ) 4 2 2 + (5 3 ) 17 5 1 (mod 31) 1 4 2 2 + 1 17 5 1 (mod 31) 4 + 5 (mod 31) 9 (mod 31) and so the remainder is 9.
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MATH 135, Midterm #2 Solutions Page 2 of 5 (d) Determine the integer a with 0 a < 7 so that [ a ] = [5] - 1 in Z 7 . [2] Solution a = 3 Reason: We want to solve [5][ a ] = [1] in Z 7 , or equivalently 5 a 1 (mod 7). The integer
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This note was uploaded on 10/21/2010 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

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Midterm2Solutions - Faculty of Mathematics University of...

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