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# 7 - Math 135 Assignment 7 Solutions Winter 2009 1(a Using...

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Math 135 Assignment 7 Solutions Winter 2009 1. (a) Using the Extended Euclidean Algorithm, we can find the gcd(1653,2000) and the required inte- gers x and y such that 1653 x + 2000 y = 77. x column y column r column 0 1 2000 1 0 1653 - 1 1 347 5 - 4 265 - 6 5 82 23 - 19 19 - 98 81 6 317 - 262 1 Thus, gcd(1653 , 2000) = 1 and since 1 | 77 the congruence has solutions. A particular solution of 1653 x 1(mod 2000) is x 0 = 317. Thus, a particular solution of 1653 x 77(mod 2000) is x 0 = 317 × 77 = 24409. Since 24409 409(mod 2000), the complete solution is x 409(mod 2000). (b) Since 2 | 1492 and 2 | 2000, 2 | gcd(1492 , 2000). But 2 77, and thus, the congruence has no solution. (c) Using the table approach and working modulo 12: x 0 1 2 3 4 5 6 7 8 9 10 11 x 2 - 4 x 0 9 8 9 0 5 0 9 8 9 0 5 Thus, the complete solution is x 0 , 4 , 6 , 10(mod 12). (d) Checking x 0, we have 5 2(mod 11). Thus, 11 x and using Fermat’s Little Theorem we have x 10 1(mod 11) and x 11 x (mod 11). The given congruence now simplifies to x + 3 + 5 2(mod 11) or x ≡ - 6(mod 11), or x 5(mod 11). 2. (a) If x = a , we get a 2 - ab ( a - b ) a (mod p ). Thus, x = a is a solution.

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7 - Math 135 Assignment 7 Solutions Winter 2009 1(a Using...

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