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Unformatted text preview: MATH 135 Algebra, Solutions to Assignment 9 1: Solve the following pairs of congruences. (a) x 5 (mod 7) x 8 (mod 15) Solution: We have x 5 (mod 7) when x { , 2 , 5 , 12 , 19 , 26 , 33 , 40 , 47 , 54 , 61 , 68 , } , and we have x 8 (mod 15) when x { , 7 , 8 , 23 , 38 , 53 , 68 , } . Thus one solution is x = 68 and so, by the Chinese Remainder Theorem, the general solution is x 68 (mod 105). (b) x 45 (mod 84) x 61 (mod 115) Solution: For x to be a solution, we need x = 45 + 84 r and x = 61 + 115 s for some integers r and s , so we need 45 + 8 r = 61 + 115 s , that is 84 r 115 s = 16. The Euclidean Algorithm gives 115 = 1 84 + 31 , 84 = 2 31 + 22 , 31 = 1 22 + 9 , 22 = 2 9 + 4 , 9 = 2 4 + 1 so we have gcd(84 , 115) = 1, and the BackSubstitution gives 1 , 2 , 5 , 7 , 19 , 26 and so we have (84)( 26) (115)( 19) = 1. Multiply both sides by 16 to get (84)( 416) (115)( 304) = 16. Thus one solution to the equation 84 r 115 s = 16 is given by ( r,s ) = ( 416 , 304), and by the Linear Diophantine Equation Theorem, the general solution is ( r,s ) = ( 416 , 304)+ k (115 , 84), k Z , so we have r  416 44 (mod 115). Thus one solution to the given pair of congruences is x = 45+84 r = 45+84 44 = 3741. Note that 84 115 = 9660, so by the Chinese Remainder Theorem, the general solution to the given pair of congruences is x 3741 (mod 9660) . 2: Solve the following pairs of congruences. (a) 15 x 4 (mod 26) 24 x 6 (mod 63) Solution: First we solve the linear congruence 15 x = 4 (mod 26). By inspection, one solution is x = 2 and we have gcd(15 , 26) = 1, and so by the Linear Congruence Theorem, the general solution is x 2 (mod 26). Next we solve 24 x 6 (mod 63). We need 24 x + 63 y = 6. The Euclidean Algorithm gives 63 = 2 24 + 15 , 24 = 1 15 + 9 , 15 = 1 9 + 6 , 9 = 1 6 + 3 , 6 = 2 3 + 0 so we have gcd(24 , 63) = 3, and then BackSubstitution gives 1 , 1 , 2 , 3 , 8 so we have (24)(8) + (63)( 3) = 3. Multiply both sides by 2 to get (24)(16) + (63)( 6) = 6. Thus one solution is x = 16. Note that 63 3 = 21, so by the Linear Congruence Theorem, the general solution to the congruence 24 x 6 (mod 63) is x 16 (mod 21). Thus the original pair of congruences is equivalent to the pair of congruences x 2 (mod 26) x 16 (mod 21) . For x to be a solution we need x = 2 + 26 r and x = 16 + 21 s for some integers r and s , so we must have 2 + 26 r = 16 + 21 s , that is 26 r 21 s = 14. The Euclidean Algorithm gives 26 = 1 21 + 5, 21 = 4 5 + 1 so we have gcd(26 , 21) = 1, and then BackSubstitution gives 1, 4, 5, so we have (26)(...
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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.
 Fall '08
 ANDREWCHILDS
 Algebra, Congruence

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