Discrete Mathematics with Graph Theory (3rd Edition) 105

# Discrete Mathematics with Graph Theory (3rd Edition) 105 -...

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Chapter 4 (c) [BB] AIR corresponds to the number 10918. E = M 5 (mod 391) = 259. (d) BYE corresponds to the number 22505. E = M 5 (mod 391) = 97. (e) NOW corresponds to the number 141523. E = M5 (mod 391) = 104. 24. Since 1 = 1(5) -1(4) and 1 = 5(5) - 4(6) we have a = 1, b = 5 in the notation of the text. 103 (a) [BB] E a (mod p) is 31 1 (mod 5) = 1; Eb (mod q) is 31 5 (mod 7) = 5. To solve M == 1 (mod 5), M == 5 (mod 7), we write 1 = 10(5) - 7(7) and get M = 5(10)(5) - 1(7)(7) = 201 == 26 (mod 35). So the message was the letter Z. (b) Ea (mod p) is 241 (mod 5) = 4; Eb (mod q) is 24 5 (mod 7) = 5. To solve M == 4 (mod 5), M == 5 (mod 7), we write 1 = 10(5) - 7(7) and get M = 5(10)(5) - 4(7)(7) = 54 == 19 (mod 35). So the message was the letter S. (c) [BB] E a (mod p) is 7 1 (mod 5) = 2; Eb (mod q) is 7 5 (mod 7) = O. To solve M == 2 (mod 5), M == 0 (mod 7), we write 1 = 10(5) - 7(7) and get M = 0(10)(5) - 2(7)(7) = -98 == 7 (mod 35). So the message was the letter G. (d) Ea (mod p) is 111 (mod 5) = 1; Eb (mod q) is 11 5 (mod 7) = 2. To solve M == 1 (mod 5), M == 2 (mod 7), we write 1 = 10(5) - 7(7) and get M = 2(10)(5) - 1(7)(7) = 51 == 16 (mod 35). So the message was the letter
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