#14 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

Given that: The ciphertext message produced by the RSA algorithm with key n,k=2573,1013 is 0464 1472 0636 1262 2111 The Euclidean algorithm yields 1013⋅17≡1mod2573. To determine the original message.If (n,k) is the public key then the ciphertext C for a plaintext P is given by, C=Pkmodn The original plaintext P from the ciphertext can be calculated as: P=Cdmodn Here d can be calculated by k⋅d≡1modϕn using the Euclidean algorithm.Here we have, n,k=2573,1013 Compare k⋅d≡1modϕn with 1013⋅17≡1mod2573 we get d=17. So, the plaintext using the ciphertext C can be calculated as: P=C17mod2573P=046417mod2573P=1704P=147217mod2573P=1511P=063617mod2573P=2426P=126217mod2573P=1314 So, the original message is 1704 1511 2426 1314The original message is 1704 1511 2426 1314