A In a publickey system using RSA you intercept the plaintex

A. In a public-key system using RSA, you intercept the plaintext M = 7 to be sent to a user. Assume p = 11, q = 13, e = 11. What is the Cypertext C? Show your calculations.

Solution

As we know that

Ciphertext (C) = M^emod n

C = 7^11 mod (p * q)

C = 7^11 mod (11 * 13) = 7^11 mod (143)

C = [ (7^3 mod 143) * (7 ^ 3 mod 143) * (7 ^ 4 mod 143) ] mod 143

C = [ 57 *57 * 113* 7 ] mod 143

C = [ 2569959 ] mod 143

C = 106

Hence Cipher Text is 106