In RSA, suppose p = 3, q = 11, e = 7, and d = 3. Given the message m = 5, show the encryption process to obtain the ciphertext c from m and the decryption process to derive m back from c.

For a block cipher with encryption Ek() and decryption DK(), the equation for the CBC mode of operation is: Cj = EK(Pj e Cj-1). What is the equation for decrypting C?

Suppose that Alice and Bob communicate using ElGamal cipher and f (p. 9. Z) is common public values. Bob generates his private key d ER Z and then computes the corresponding and public public key y=g" (mod p). To save time, Bob uses the same number r each time he encrypts a plaintext message m (ie., r is a fixed nonce of Bob, and it is not randomly generated each time encryption is performed). Assume that Alice compute the ciphertext for the message m as (cc) = (g mod p, mxy mod p). and for the message m as (1,2)=(g" mod p, xy' mod p). Show how an adversary who possesses a plaintext-ciphertext pair (m. (c.ca)) can decrypt (1, 2) without knowing the private key d of Bob.

Suppose your RSA modulus is n = 55 = (5)(11) and your encryption exponent is e = 3. a) Find the decryption modulus d. b) Assume that gcd(m,55)=11. Show that if c ≡ m3 ( mod 55) is the ciphertext, then the plaintext is m ≡ c d ( mod 55)