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Discrete Mathematics with Graph Theory (3rd Edition) 81

Discrete Mathematics with Graph Theory (3rd Edition) 81 -...

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Section 4.2 (b) We have 260 1 154 0 106 1 48 -1 10 3 8 -13 2 16 0 1 -1 2 -5 22 -27 and so gcd(154, 260) = 2 = 16(260) + (-27)(154). Multiplying by 2, 4 = 32(260) + (-54)(154): x = -54, y = 32. 79 (c) [BB] Consider the possibility that 154x + 260y = 3 for certain integers x and y. The left side of this equation is even while the right side is odd, a contradiction. (d) Suppose we have 196x + 260y = 14 for some integers x and y. Since the left side is divisible by 4, the right side would have to be divisible by 4, which is not the case. 18. (a) [BB] We know that gcd(x, y) I x and gcd(x, y) I y and so gcd(x, y) I (mx + ny) (Proposi- tion 4.2.3). (b) Yes, it is true. Assume gcd(x, y) I d so that d = i(gcd(x, y)) for some integer i. By Theo- rem 4.2.9, there are integers s and t such that gcd(x, y) = sx + ty. Hence, d = isx + tty, and m = is, n = it will work. 19. (a) Take x = -6, y = 9, for example. (b) Suppose7x+5y = 3. Since 7(-6)+5(9) = 3aswell,subtractinggives 7(x+6)+5(y-9) = O. Since 717(x + 6), so also 715(y - 9). Since the integers 5 and 7 are relatively prime, we must have 71 y - 9, so y - 9 = 7k for some integer k. Thus y = 9 + 7k and 7(x + 6) = -5(y - 9) = -5(7k), so x + 6 = -5k and x = -6 - 5k.
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