MATH 406 Midterm: MATH406 HERB-R FALL2006 0101 MID SOL
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15 Feb 2019
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Math 406 - exam 1 - oct. 6, 2006- solutions. 4 = 14 10, 2 = 10 2 4 = 10 2(14 10) = ( 2)(14) + 3(10). Thus one solution is x0 = 2, y0 = 3. Now all solutions are given by x = 2 + (10/2)k = 2 + 5k, y = 3 (14/2)k = 3 7k where k is any integer. (a) gcd(18, 20) = 2 and 2|12. 3. x0 = ( 1)(12)/2 = 6 and x = 6+k(20)/2 = 6+10k, k = 0. 1. Since 6 14 (mod 20), a complete list of incongruent solutions is x 4, 14 (mod 20). (b) gcd(24, 30) = 6 and 6 6 | 3. Thus there are no solutions. (c) x 0 (mod 5) is not a solution. Little theorem x4 1 (mod 5), so that x30 x2 (mod 5). But mod 5, 12 1, 22 .
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