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Unformatted text preview: Math 55: Discrete Mathematics, Fall 2008 Homework 4 Solutions 3.7: 32. First of all, factor 1729 = 7 13 19. Now notice that for each of the prime factors p , p 1 divides 1728. Hence if x 6 0 (mod 1729), then Fermats theorem implies that x 1728 1 (mod 7), x 1728 1 (mod 13), and x 1728 1 (mod 19). By the Chinese Remainder Theorem, it follows that x 1728 1 (mod 1729), that is, 1729 is a Carmichael number. [This problem was also done in lecture.] 46. The letter pairs are represented by the integers 120, 2001, 311. Encrypting these with E ( x ) = x 13 (mod 2537), we get 286, 798, 425. (A) The decryption exponent is 13 1 (mod 42 58), or d = 937. Computing 286 937 (mod 2537) gives back 120, as expected. *60. The solutions are x 4 , 11 , 31 , 46 (mod 105). To find them, observe that modulo each prime, the solutions are x 4. Thus x 1 (mod 3), x 1 (mod 5), x 3 (mod 7). For each of the eight possible combinations of signs, solve the three simultaneous congruences using the Chinese Remainder Theorem to...
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 Fall '08
 STRAIN
 Math

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