Discrete Mathematics with Graph Theory (3rd Edition) 103

# Discrete Mathematics with Graph Theory (3rd Edition) 103 -...

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Section 4.5 (d) [BB] We wish to solve the system of congruences x 0 (mod 4) x - 2 (mod 3) x 10 (mod 25). 101 Solving the first two congruences, we obtain x == 8 (mod 12). Since 1(25) - 2(12) = 1, x = 8(25) - 10(2)(12) = -40 == 260 (mod 300). (e) We wish to solve the system of congruences x 3 (mod 4) x - 2 (mod 3) x 24 (mod 25). Solving the first two congruences, we obtain x == 11 (mod 12). Since 1(25) - 2(12) = 1, x = 11(25) - 24(2)(12) = -301 == 299 (mod 300). 20. [BB] Call the number x. Then x == 4 (mod 12) and x == 15 (mod 25). By the Chinese Remainder Theorem, this system has a solution which is unique mod 12(25) = 300. There is exactly one such number in the interval 200 < x < 500 so the given congruences determine the number uniquely. Since 1(25) - 2(12) = 1, the method suggested in the text gives x == 4(1)(25) - 15(2)(12) = -260 == 40 (mod 300). The number was 340. 21. (a) [BB] ab == 3(3) = 1 (mod 4) ab == 0(8) = 0 (mod 9) ab == 4(18) = 22 (mod 25). Since 1 = 9 - 2(4), we get ab == 1(9) - 0(2)(4) = 9 (mod 36) as the solution to the first two congruences. Then, since 1
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