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Unformatted text preview: Math 347 C1 HOUR EXAM II 21 July 2009 SOLUTIONS 1. Consider the equation: 5 x + 2 y = 37 . a) Find integers x , y , which satisfy the equation. b) Find integers x , y , with y positive, which satisfy the equation. (If your answer to part a) has y > 0, then you are done.) SOLUTION a) Since 5 and 2 are relatively prime, there are integers m and n such that 5 m + 2 n = 1. By inspection, m = 1 and n = 2 work, i. e. (5 1) + (2 ( 2)) = 1 . Then we can multliply this equation by 37 to get (5 37) + (2 ( 2)37) = 37 , that is (5 37) + (2 ( 74)) = 37 . Thus, x = 37 and y = 74 is a solution. b) If we replace y = 74 by y = ( 74) + (5 37) and x = 37 by x = 37 (2 37), then we have (5(37 2(37))+(2( 74+(5(37)) = 37 (We have added and subtracted 10 37.) or 5( 37) + 2(111) = 37 . Thus x = 37 and y = 111. 2. a) Find a polynomial p ( x ) of degree 3 with integral coefficients, such that p ( x ) 0 (mod 3), for all x Z . b) Is there a polynomial q ( x ) of degree = 1 or 2 with integral coefficients, such that q ( x ) 0 (mod 3), for all x Z ? Explain....
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 Spring '09
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 Math, Integers

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