(RSA encryption) Let n = 5 17 = 85 be the modulus of a (very modest)...

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(RSA encryption) Let n = 5 17 = 85 be the modulus of a (very modest) RSA public key encryption and d = 5 the decryption key. Since 85 is in between 25 and 2525, we can only encode one letter (with a two-digit representation) at a time. a) (10 pts) Use the decryption function M = Cd mod n = C³ mod 91 to decipher the 5 encrypted message 68-44-00-52-59. b) (10 pts) What is the encryption key? Use that to write the encryption function. (RSA encryption) Let n = 5 17 = 85 be the modulus of a (very modest) RSA public key encryption and d = 5 the decryption key. Since 85 is in between 25 and 2525, we can only encode one letter (with a two-digit representation) at a time. a) (10 pts) Use the decryption function M = Cd mod n = C³ mod 91 to decipher the 5 encrypted message 68-44-00-52-59. b) (10 pts) What is the encryption key? Use that to write the encryption function.

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a To decipher the encrypted message using RSA you have the modulus n 517 and the decryption key d 5 ... View the full answer