Suppose that both plaintexts and ciphertexts consist of trigraph message units,

Question

Suppose that both plaintexts and ciphertexts consist of trigraph message units, but while plaintexts are written in the 27-letter alphabet (consisting of A-Z and blank=26), ciphertexts are written in the 28- letter alphabet obtained by adding the symbol "/" (with numerical equivalent 27) to the 27-letter alphabet. We require that each user A choose n_A between 27^3 = 19683 and 28^3 = 21952, so that a plaintext trigraph in the 27-letter alphabet corresponds to a residue P modulo n_A, and then C = P^eA mod n_A corresponds to a ciphertext trigraph in the 28-letter alphabet. (a) If your deciphering key is K_D = (n, d) = (21583, 20787), decipher the message "YSNAUOZHXXH " (one blank at the end). (b) If in part (a) you know that phi(n) = 21280, find (i) e = d^-1 mod phi(n), and (ii) the factorization of n.

Explanation / Answer

n = 21583

n = p*q = 113 * 191 = 21583

(n) = (p-1)(q-1) = 112 * 190 = 21280

e = d-1 mod (n) = 45

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