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# key5fa02 - COT3100C-01 Fall 2002 S Lang Key to...

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COT3100C-01, Fall 2002 S. Lang Key to Assignment #5 (40 pts.) 11/14/2002 1. (10 pts.) Compute GCD(1027, 373) using Euclid’s algorithm, and find two integers t and u such that 1027 t + 373 u = GCD(1027, 373). The following steps show how the extended Euclid’s GCD algorithm is applied: 1027 = 373 2 + 281 --- (1) 373 = 281 1 + 92 --- (2) 281 = 92 3 + 5 --- (3) 92 = 5 18 + 2 --- (4) 5 = 2 2 + 1 --- (5) 2 = 1 2 + 0 Thus, GCD(1027, 373) = 1 = 5 – 2 2 , using (5) = 5 – (92 – 5 18) 2, using (4) = 92 (–2) + 5 (1+36) = 92 (–2) + 5 37 = 92 (–2) + (281 – 92 3) 37, using (3) = 281 37 + 92 (–2 – 3 37) = 281 37 + 92 (–113) = 281 37 + (373 – 281 1) (–113), using (2) = 373 (–113) + 281 (37 + 113) = 373 (–113) + 281 150 = 373 (–113) + (1027 – 373 2) 150, using (1) = 1027 150 + 373 (–113 – 2 150) = 1027 150 + 373 (–413) Therefore, t = 150, and u = –413, and 1027 t + 373 u = GCD(1027, 373) = 1. 2. (10 pts.) Suppose a , b , and c are positive integers. Prove that if GCD( a , b ) = GCD( a , c ) = 1, then GCD( a , bc ) = 1. Proof: We use proof by contradiction. That is, suppose GCD( a , bc ) = n > 1 --- (1). We want to prove this leads to a contradiction. From (1), n must have a prime factor p according to the fundamental theorem of arithmetic. Thus, p | a --- (2) and p | bc --- (3) because p | n and n = GCD( a , bc ). Note that (3) implies

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key5fa02 - COT3100C-01 Fall 2002 S Lang Key to...

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