# soln5 - MATH 135 Algebra Solutions to Assignment 5 1 Solve...

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Unformatted text preview: MATH 135 Algebra, Solutions to Assignment 5 1: Solve each of the following linear diophantine equations. (a) 42 x + 30 y = 24 Solution: The Euclidean Algorithm gives 42 = 30 · 1 + 12 . 30 = 12 · 2 + 6 , 12 = 6 · 2 + 0 so we have gcd(42 , 30) = 6. Back-Substitution then gives 1 ,- 2 , 3 so we have 42(- 2)+30(3) = 6. Note that 24 6 = 4, and multiplying both sides by 4 gives 42(- 8)+30(12) = 24, and so one solution is ( x,y ) = (- 8 , 12). Note that 42 6 = 7 and 30 6 = 5, so by the Linear Diophantine Equation Theorem, the general solution is ( x,y ) = (- 8 , 12) + k (- 5 , 7) , k ∈ Z . (b) 231 x + 792 y = 513 Solution: The Euclidean Algorithm gives 792 = 3 · 231 + 99 , 231 = 2 · 99 + 33 , 99 = 3 · 33 + 0 and so gcd(231 , 792) = 33. Back-Substitution gives 1 ,- 2 , 7 and so 231(7) + 792(- 2) = 33. Note that 513 = 33 · 15 + 18, so 33 does not divide 513, and hence there is no solution (by Proposition 2.11(ii) or by the Linear Diophantine Equation Theorem). (c) 385 x- 1183 y = 294 Solution: The Euclidean Algorithm gives 1183 = 3 · 385 + 28 , 385 = 13 · 28 + 21 , 28 = 1 · 21 + 7 , 21 = 3 · 7 + 0 and so gcd(385 , 1183) = 7. Back Substitution gives 1 ,- 1 , 14 ,- 43 and so we have 385(- 43) + 1183(14) = 7. Note that 294 7 = 42, and multiplying both sides by 42 gives 385(- 1806)- 1183(- 588) = 294. Thus one solution is ( x,y ) = (- 1806 ,- 588). Note that 385 7 = 55 and 1183 7 = 169, and so by the Linear Diophantine Equation Theorem, the general solution is ( x,y ) = (- 1806 ,- 588) + k (169 , 55) , k ∈ Z 2: (a) Find all non-negative solutions to the diophantine equation 483 x + 336 y = 9513. Solution: The Euclidean Algorithm gives 483 = 1 · 336 + 147 , 336 = 2 · 147 + 42 , 147 = 3 · 42 + 21 , 42 = 2 · 21 + 0 so gcd(483 , 336) = 21. Back-Substitution gives 1 ,- 3 , 7 ,- 10 so we have 483(7)+336(- 10) = 21. Note that 9513 21 = 453 and multiplying both sides of the equation by 453 gives 483(3171) + 336(- 4530) = 9513. Thus one solution is ( x,y ) = (3171 ,- 4530). Note that 483 21 = 23 and 336 21 = 16, so by the Linear Diophantine Equation Theorem, the general solution is ( x,y ) = (3171 ,- 4530) + k (- 16 , 23) , k ∈ Z . Note that x ≥ 0 = ⇒ 3171- 16 k ≥ 0 = ⇒ 16 k ≤ 3171 = ⇒ k ≤ 3171 16 = 198 y ≥ 0 = ⇒ - 4530 + 23 k ≥ 0 = ⇒ 23 k ≥ 4530 = ⇒ k ≥ 4530 23 = 197 so we obtain non-negative solutions when k = 197 and 198. The solutions are ( x,y ) = (19 , 1) , (3 , 24) . (b) Find all pairs of integers ( x,y ) with x ≥ 1000, y ≤ 1000 such that 726 x- 1578 y = 324. Solution: The Euclidean Algorithm gives 1578 = 2 · 726 + 126 , 726 = 5 · 126 + 96 , 126 = 1 · 96 + 30 , 96 = 3 · 30 + 6 , 30 = 5 · 6 + 0 so we have gcd(726 , 1578) = 6. Back-Substitution gives 1 ,- 3 , 4 ,- 23 , 50 so we have 726(50)- 1578(23) = 6. Note that 324 6 = 54 and multiplying both sides of the equation by 54...
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## This note was uploaded on 05/04/2010 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

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soln5 - MATH 135 Algebra Solutions to Assignment 5 1 Solve...

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