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Unformatted text preview: MATH 135 Assignment 2, Winter 2010 Due 8:30am on Wednesday January 20, 2010 in the drop boxes opposite the Tutorial Centre, MC 4066. Instructions: First attempt questions 12 and use page 2 to check your answers (do not turn these in). Hand in the solutions for questions 36 by the due date (total 38 marks). Question 7 is for extra credit, due the same time as the rest. A solution to an LDE is positive if the solutions to x and y are positive simultaneously, and nonnegative if theyre both nonnegative. Additional practice problems are 39, 43, 53, 57, 63, 65 on p51 of the textbook. Question 1. Practice question on the EEA and the gcd characterization theorem [0 marks] Prove that gcd( ab, c ) = 1 if and only if gcd( a, c ) = gcd( b, c ) = 1. Question 2. Practice questions on linear Diophantine equations [0 marks] For each linear Diophantine equation, explain whether there is a solution or not. Write down a complete solution if one exists. (a) 14 x + 91 y = 200 (b) 331 x + 249 y = 1 (c) 331 x 249 y = 3 Question 3. Linear Diophantine equation with constraints [Parts (a) and (b) carry 6 and 8 marks respectively, 14 marks total]. (a) Determine all solutions to the linear Diophantine equation 57 x + 135 y = 8400. (b) Find all the nonnegative solutions of the above that also satisfy the inequality x + 2 y 140. Answer: (a) Applying EEA to 135 and 57: 1 135 1 57 1 2 21 2 5 15 3 7 6 8 19 3 2 135 = 57 2 + 21 2 57 = 21 2 + 15 1 21 = 15 1 + 6 2 15 = 6 2 + 3 2 6 = 3 2 + 0 Noting the ordering of 57 and 135 in the table, we conclude gcd(57 , 135) = 3 = 57(19) + 135( 8). Multiplying the above by 2800, we get 8400 = 57(53200) + 135( 22400). So x = 53200 , y = 22400 is a particular solution to the given LDE. The complete solution is x = 53200 + (135 / 3) n = 53200 + 45 n, y = 22400 (57 / 3) n = 22400 19 n n Z . 1 Answer: (b) We want to find the values of n so that x 0, y 0, and x + 2 y 140....
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 Spring '08
 ANDREWCHILDS
 Math

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