Hint to the solution

3. Suppose that both plaintexts and ciphertexts consist of trigraph mes sage 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 "V" (with numerical equivalent 27) to the 27-letter alphabet. We require that each user A choose nA between 27 = 19683 and 283 = 21952, so that a plaintext trigraph in the 27-letter alphabet corresponds to a residue P modulo nAand then C = peA mod nA corresponds to a ciphertext trigraph , in the 28-letter alphabet. (a) If your deciphering key is KD = ) (,, (n, d= 2158320787)decipher the message YSNAUOZHXXH (one blank at the end). (b) If in part (a) you know that pn) = 21280, find (i)e = dimod Plm), and (ii) the factorization of n. , or it interer 23360947609 is a particularly bad choice