Exam1Sol - COT 5937 Summer 2004 Exam#1 Solutions 1 The...

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COT 5937 Summer 2004 Exam #1 Solutions 1) The given information leads to setting up the two following equations: f(17) = 17a + b = 40 (mod 52) f(50) = 50a + b = 19 (mod 52) Subtracting the top equation from the bottom we get 33a = -21 (mod 52) Since gcd(52, 33) = 1, we can use the Extended Euclidean Algorithm to find 33 -1 mod 52. 52 = 1x33 + 19 33 = 1x19 + 14 19 = 1x14 + 5 14 = 2x5 + 4 5 = 1x4 + 1 5 - 1x4 = 1 5 - (14 - 2x5) = 1 3x5 - 14 = 1 3(19 - 14) - 14 = 1 3x19 - 4x14 = 1 3x19 - 4(33 - 19) = 1 7x19 - 4x33 = 1 7(52 - 33) - 4x33 = 1 7x52 - 11x33 = 1 Evaluating this equation mod 52, we find that (-11)x33 = 1 (mod 52). Thus, 33 -1 = -11 (mod 52). Now, solve for a: 33a = -21 (mod 52) (-11)(33)a = (-11)(-21) (mod 52) a = 231 (mod 52) a = 23 (mod 52) Now, to solve for b, plug into the first original equation: f(17) = 17a + b = 17(23) + b = 40 (mod 52) 391 + b = 40 (mod 52) 27 + b = 40 (mod 52) b = 13 (mod 52)
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2) Here is the 5x5 grid used for encryption: Z X W V U T S Q O N M K H G E D C B R J I F Y A L P / . Here is the plaintext split into pairs and the ciphertexts for each pair: Plain GO TO AF AS TF OX OD WI ND OW Cipher OV NQ YP YQ ZD SV TR UB TI/TJ QV 3)
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