hw04sol

# hw04sol - Math 55 Discrete Mathematics Fall 2008 Homework 4...

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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 Â· 13 Â· 19. Now notice that for each of the prime factors p , p- 1 divides 1728. Hence if x 6â‰¡ 0 (mod 1729), then Fermatâ€™s theorem implies that x 1728 â‰¡ 1 (mod 7), x 1728 â‰¡ 1 (mod 13), and x 1728 â‰¡ 1 (mod 19). By the Chinese Remainder Theorem, it follows that x 1728 â‰¡ 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 Â· 58), or d = 937. Computing 286 937 (mod 2537) gives back 120, as expected. *60. The solutions are x â‰¡ Â± 4 , Â± 11 , Â± 31 , Â± 46 (mod 105). To find them, observe that modulo each prime, the solutions are x â‰¡ Â± 4. Thus x â‰¡ Â± 1 (mod 3), x â‰¡ Â± 1 (mod 5), x â‰¡ Â± 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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## This note was uploaded on 08/16/2010 for the course MATH 55 taught by Professor Strain during the Fall '08 term at Berkeley.

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