Discrete Mathematics with Graph Theory (3rd Edition) 95

# Discrete Mathematics with Graph Theory (3rd Edition) 95 -...

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Section 4.4 93 vi. No x exists. The values of 4x 2 + 3x + 7 (mod 5) are 2, 3 and 4. 11. (a) [BB] Multiplying the second congruence by 2 gives 2x == 0 (mod 6) and, hence, x == 0 or x == 3 (mod 6). If x == 0, then the first congruence says y == 1 and it is easily checked that x = 0, y = 1 satisfies both congruences. If x == 3, then the first congruence says y == 1, but these values for x and y do not satisfy the second congruence. There is just one solution mod 6; namely, x = 0, y = 1. (b) [BB] Subtracting the first congruence from the second gives 3x == -2 == 7 (mod 9). But the values of 3x mod 9 are just 0, 3 and 6. There is no solution. (c) Subtracting the first congruence from the second gives 3x == -2 == 6 (mod 8) and, hence, x == 2. The first congruence then says 2 + 5y == 3; that is, 5y == 1 (mod 8). So y == 5 (mod 8). One checks easily that x = 2, y = 5 is a solution to the given congruences. (d) Multiply the first congruence by 2 to get
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## This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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