MATH135_F11_Assignment_4_Solns

MATH135_F11_Assignment_4_Solns - 1 MATH 135 F 2011:...

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1 MATH 135 F 2011: Assignment 4 Due: 8:30 AM, Wednesday, 2011 Oct. 26 in the dropboxes outside MC 4066 Write your answers in the space provided. If you wish to typeset your solutions, use the solution template posted on the course web site. Typesetting is done in L A T E X. A very good online all-purpose L A T E Xmanual is the L A T E Xwikibook at http://en.wikibooks.org/wiki/LaTeX . Links to installations for various operating systems are also included there. Stephen Carr is IST’s LaTeX goto person, and his introductory guide is at http://www.ist.uwaterloo.ca/ew/saw/latex/latex_getstarted.pdf . L A T E Xis totally voluntary. Cite any proposition or definition you use. Family Name: Solutions First Name: I.D. Number: Section: Mark: (For the marker only.) If you used any references beyond the course text and lectures (such as other texts, discussions with colleagues or online resources), indicate this information in the space below. If you did not use any aids, state this in the space provided. 1. What is the remainder when 2 271 3 314 is divided by 7? Provide justification for your work. First, observe that 2 3 1 (mod 7) and 3 3 ≡ - 1 (mod 7) and so by the proposition on the Properties of Congruence, 2 271 3 314 (2 3 ) 90 2 1 (3 3 ) 104 3 2 (1) 90 2 1 ( - 1) 104 3 2 2 · 9 18 4 (mod 7) This, by the proposition Congruent Iff Same Remainder, 2 271 3 314 has remainder 4 when divided by 7.
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2 2. For each linear congruence, determine the complete solution, if a solution exists. (a) 3 x 11 (mod 18) Since gcd(3 , 18) = 3 and 3 - 11, this linear congruence has no solution by LCT 1. (b) 4 x 5 (mod 21) Since gcd(4 , 21) = 1 and 1 | 5, this linear congruence has exactly one solution modulo 21 by LCT 1. By inspection, x 17 (mod 21). (c) 36
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This note was uploaded on 12/31/2011 for the course MATH 135 taught by Professor Andrewchilds during the Spring '08 term at Waterloo.

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MATH135_F11_Assignment_4_Solns - 1 MATH 135 F 2011:...

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