#13 please in thatollowedph cipheINTRODUCTION TO CRYPTOGRAPHY60I1. Encrypt the message GOOD CHOICE using an exponential cipher with modulus ,2609 and exponent k = 7.12. The ciphertext obtained from an exponential cipher with modulus p = 2551 and enci-phering exponent k = 43 is= d1518 2175 1249 0823 2407Determine the plaintext message.13. Encrypt the plaintext message GOLD MEDAL using the RSA algorithm with key(2561,3).14. The ciphertext message produced by the RSA algorithm with key (n, k) = (2573, 1013) is046414720636 1262 2111Determine the original message. [Hint: The Euclidean algorithm yields 1013 · 17 =1 (mod 2573).]15. Decrypt the ciphertext1030 1511 0744 1237 1719that was encrypted using the RSA algorithm with key (n, k) = (2623, 869). [Hint: Therecovery exponent is j = 29.]%3D10.2 THE KNAPSACK CRYPTOSYSTEMA public-key cryptosystem also can be based on the classic problem in combinatoricsknown as the knapsack problem, or the subset sum problem. This problem may bestated as follows: Given a knapsack of volume V and n items of various volumes42, ..., an, can a subset of these items be found that will completely fill theknapsack? There is an alternative formulation: For positive integers a1, a2, . .., Anand a sum V , solve the equationUxup +...+ + D = /Where xị = 0 or 1 for i = 1, 2, . . . , n.the onght be no solution, or more than one solution, to the problem, dependngknapsack problemis not solvable; but3D%3=has two distinct solutions

According to the given information and you mention the question number 13 :- Encrypt plaintext message GOLD MEDAL Using RSA algorithm with key 2561.3M=0614 1103 2612 0403 0011 using the encryption key (2561.3) the message can be broken into blocks of length 4. then, module 2561 06143 = 2456 1103 = 0687 26123 = 1632 04033 = 2059 00113 = 1131 So that the ciphertext message is 2456 0687 1632 2059 1331