Midterm2Solutions

# Midterm2Solutions - Faculty of Mathematics University of...

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Unformatted text preview: Faculty of Mathematics University of Waterloo MATH 135 MIDTERM EXAM #2 Fall 2007 Monday 12 November 2007 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) Convert (2345) 6 to base 10. [3] Answer: (2345) 5 = 569 Solution By definition, (2345) 6 = 2(6 3 ) + 3(6 2 ) + 4(6) + 5 = 2(216) + 3(36) + 4(6) + 5 = 569 so (2345) 6 = (569) 10 . (b) Convert (2007) 10 to base 12, using A and B to represent the digits 10 and 11, respectively. [3] Answer: (2007) 10 = (11 B 3) 12 Solution Using the conversion algorithm, 2007 = 167(12) + 3 167 = 13(12) + 11 13 = 1(12) + 1 1 = 0(12) + 1 Therefore, (2007) 10 = (11 B 3) 12 , since B represents the digit 11. (c) Determine the remainder when 2 34 56 52 + 3 19 is divided by 17. [3] Answer: 11 Solution Since 17 is prime and none of 2, 56 or 3 is divisible by 17, then 2 16 56 16 3 16 1 (mod 17) by Fermats Little Theorem. MATH 135, Midterm #2 Solutions Page 2 of 5 Therefore, 2 34 56 52 + 3 19 (2 16 ) 2 2 2 (56 16 ) 3 56 4 + (3 16 )3 3 (mod 17) 1 2 (4)1 3 5 4 + 1(27) (mod 17) (since 56 5 (mod 17)) 4(625) + 27 (mod 17) 2527 (mod 17) 11 (mod 17) Therefore, the remainder is 11. (d) Determine the number of congruence classes in Z 18 that are solutions to the equation [3] [12][ x ] = [5]....
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Midterm2Solutions - Faculty of Mathematics University of...

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