Question: # 2 . With modulus p = 4 1 and unknown encryption key e , modular exponentiation produces the ciphertext 3 5 1 3 3

#2. With modulus p=41 and unknown encryption key e, modular exponentiation
produces the ciphertext
35133126310516111511050703
Cryptanalyze the above cipher, given that the ciphertext block 11 corresponds to
plaintext M; what is the plaintext message? Note that we don't have any quick way
to solve for e just yet. Use trial and error. (8.3)
# 2 . With modulus p = 4 1 and unknown encryption

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