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9. (RSA encryption) Let n = 17 be the modulus of a (very modest) RSA public key encryption and d = 3 the decryption key: Since 85 is in between 25 and 2525, we can only encode one letter (with two-digit representation) at a time.a. (15 pts) Use the functionM = Cd mod nto decipher the six-letter decrypted message 08 = 44 - 45 = 45 - 64 - 64.b. (5 pts) Determine the other portion of the public key in the RSA above (find the exponent e). Then write down the encryption function.
9. (RSA encryption) Let n = 17 be the modulus of a (very modest) R…
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SOLVED: ExercisesWhat is the original message encrypled using the RSA system with n = 53 61 and e = 17 ifthe encrypled message is 3185 2038 2460 25502 (To decryp first find the decryption exponent d, which is the inverse of e = [7 modulo 52 60.) Encrypt the message UPLOAD using the RSA system with n = 53 6] and e-4 translating each letter into integers and grouping together pairs of integers Encrypt the message ATTACK using the RSA system with n ande=; translating cach letter into integers and g
SOLVED: ExercisesWhat is the original message encrypled usin…
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