Discrete Mathematics with Graph Theory (3rd Edition) 107

Discrete Mathematics with Graph Theory (3rd Edition) 107 -...

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Chapter 4 105 97 1 0 17 0 1 10. (a) We use the Euclidean algorithm: 12 1 -5 5 -1 6 Thus 1 = -7(97) + 40(17). 2 3 -17 1 -7 40 (b) By part (a), 10 = -70(97) + 400(17). Mod 97, this says 400(17) (mod 97). 10, so x = 400 == 11. (a) We know the solution is unique since gcd 8,27 = 1. A patient person might use trial and error to get x = 17. Alternatively, using the method suggested by Problem 25, writing 1 = 3(27) - 10(8) shows that x = -10 == 17 (mod 27). (b) As in part (a), we know the solution is unique. Since 3(17) == 1 (mod 27), 3(15)(17) == 15 (mod 27), so x = 3(15) = 45 == 18 (mod 27). (c) We want 16 I (4x - 8), so 4 I (x - 2); that is, x - 2 = 4k for some k. The values of x (mod 16) are x = 2,6, 14. (d) There is no solution because the values of 4x (mod 16) are 0, 4, and 8. (e) We want 380 I (6x - 7), so 6x - 7 = 380k for some integer k. This says 7 = - is even, which is not true. There is no solution. (f) We want 380 I - 8), so - 8 = for some integer k. This is equivalent to 3x = 190k + 4 = 4, 194, 384, 574, 764, 954, ... so x = 128 or x = 318. 12. (a) Since gcd(6, 25) = 1, we know there exist integers a and b so that 6a + 25b = In fact, we find that 6( -4) + 25(1) = Multiplying by 13 gives 6( -52) + 25(13) = 13, so take Xo = -52 and Yo = 13.
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