# Hmk1Sol - COT 5937 2004 Summer Homework#1 Solutions 1...

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COT 5937 2004 Summer Homework #1 Solutions 1) Encryption Key: C R Y P T O G A H B D E F I K L M N Q S U V W X Z Divide plaintext into pairs: PL EA SE ME ET ME AT TH EP AR KT OM OR RO WA TX TW OP MI FI TI SR AI NI NG IW IL LB EB EH IN DT HE RE DB EN CH Encryption Process: Eg. PL, P is at the first row and L is at the first column, so we get C; L is at the 4 th row and P is at the 4 th column, so we get Q; Thus PL is converted to CQ; ME, they are at the same column, so each is converted to the character below them. M is converted to V, and E is converted to M, thus ME is converted to VM; FI, they are at the same row, so each is converted to the character at the right. F is converted to I, and I is converted to K, thus FI is converted to IK; Final Cipher Text: CQ FG MK VM KR VM BY PB IR GY SB GL GC CG YF PZ YZ HC QE IK PK MT HF QF MA FX DQ SO KG IG FQ KC GI GM KO FM PO 2) To determine the number of valid keys, note that we must be able to uniquely solve the equation y (ax + b) (mod 35) for x in terms of a, b, and y. Consequently, we find that ax (y - b) (mod 35) This equation has an unique solution if and only if a -1 mod 35 exists. As discussed in class, this occurs precisely when gcd(a, 35) = 1. There are 24 values that satisfy this. They are 1, 2, 3, 4, 6, 8, 9, 11, 12, 13, 16, 17, 18 19, 22, 23, 24, 26, 27, 29, 31, 32, 33, and 34. (There's a quicker way to determine this that we'll learn in chapter 8 of the text.) b can be any of 35 values, thus the total number of possible keys is 24x35 = 840. Assume we are given a and b, and have to determine the number of fixed points. In essence, we are looking for the number of solutions to the equation x (ax + b) mod 35 With some algebra, we find:

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x - ax b mod 35 x(1- a) b mod 35 (a - 1)x -b mod 35 At this point you'll notice that as long as (a-1)
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Hmk1Sol - COT 5937 2004 Summer Homework#1 Solutions 1...

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