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Unformatted text preview: n = 143 and a = 103 were used. Decode them to numbers and then decode them to letters using the following agreement: J = 1 ,N = 2 ,R = 3 ,H = 4 ,D = 5 ,A = 6 ,S = 7 ,Y = 8 ,T = 9 ,O = 0 Problem 4. Find the inverse of ± 1 2 3 4 5 6 7 8 9 10 7 5 6 9 10 2 1 4 8 3 ² 1 Problem 5. Calculate πσ and σπ and check if they are equal π = ± 1 2 3 4 5 6 7 7 5 6 2 1 4 3 ² σ = ± 1 2 3 4 5 6 7 3 2 6 1 7 4 5 ² 2...
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 Spring '08
 MULLEN
 Math, Cryptography, Exponentiation, Grammatical number, public key code

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