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Unformatted text preview: Assignment 8 â€“ Solutions 1. Solve the following pairs of congruences. 15 x â‰¡ 4 (mod 26) 24 x â‰¡ 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 â‰¡ 2 (mod 26). Next we solve 24 x â‰¡ 6 (mod 63). We need 24 x + 63 y = 6. The Euclidean Algorithm gives 63 = 2 Â· 24 + 15 , 24 = 1 Â· 15 + 9 , 15 = 1 Â· 9 + 6 , 9 = 1 Â· 6 + 3 , 6 = 2 Â· 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 â‰¡ 6 (mod 63) is x â‰¡ 16 (mod 21). Thus the original pair of congruences is equivalent to the pair of congruences x â‰¡ 2 (mod 26) x â‰¡ 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 Â· 21 + 5, 21 = 4 Â· 5 + 1 so we have gcd(26 , 21) = 1, and then BackSubstitution gives 1, 4, 5, so we have (26)( 4) (21)( 5) = 1. Multiply both sides by 14 to get (26)( 56) (21)( 70) = 14. Thus one solution is ( r,s ) = ( 56 , 70) and the general solution is ( r,s ) = ( 56 , 70) + k (21 , 26), k âˆˆ Z , so we have r â‰¡  56 â‰¡ 7 (mod 21). Thus one solution to the pair of congruences is7 (mod 21)....
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This note was uploaded on 07/24/2011 for the course MATH 135 taught by Professor Andrewchilds during the Winter '08 term at Waterloo.
 Winter '08
 ANDREWCHILDS
 Math, Congruence

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