A.Given p=25, q=9, and e=7, generate the public key (n,e) and the private key (n,d) using the RSA Key generation algorithm. (Use Excel for computations) b. Given Bob's public key of (85, 7) and private key of (85, 55), show how Alice can encrypt m=3 to send to Bob. Show the ciphertext.
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A.Given p=25, q=9, and e=7, generate the public key (n,e) and the private key (n,d) using the RSA Key generation
b. Given Bob's public key of (85, 7) and private key of (85, 55), show how Alice can encrypt m=3 to send to Bob. Show the ciphertext.
c. Given Bob's public key of (85, 7) and private key of (85, 55), show how Bob can decrypt the ciphertext y=2 received from Alice. Show the plaintext.
d. Using the fast exponentiation method, determine 530 mod 47. Show your work.
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- For this problem you will perform RSA encryption and decryption with p = 13, q = 17, and e = 5 (not 3!). Show your work. Compute n and d. Encrypt the message m = 14 and give the ciphertext. Perform decryption on the ciphertext obtained in part (b) to get the plaintext message.Alice and Bod have decided to use a symmetric encryption algorithm. They have some assumptions about their messages:- Messages only contain capital letters (i.e. A to Z)- The length of their shared key must be greater than or equal to the length of the plaintext- They assign each letter a number as follows: (A,0), (B,1), (C,2), (D,3),…, (Z,25)Their algorithm combines the key and the message using modular addition. The numerical values of corresponding message and key letters are added together, modulo 26. For example, if the plain text is “HELLO” and the key is “SECRET” then the encrypted message is calculated as following:Since the length of the plaintext is 5, we just need the first 5 letters of the key (i.e. “SECRE”), then for each letter, we should add corresponding letters in both the plaintext and the key modulo 26.Plaintext: H (7) E (4) L (11) L (11) O (14)Key: S (18) E (4) C (2) R (17) E(4)Cipher: Z (25) I (8) N(13) C(2) S (18) Write a program in Python, C/C++ or JavaScript to…Let p = 37 and q = 31 in the RSA set up. A. What are all the valid exponents in this case? Pick a specific value of exponents, e1. For this exponent (public key) compute the corresponding private key d1. Take message x = random value from the set of allowed values? Define this set explicitly. For this x, compute the cipher y using the public key. Decrypt this y using the private key. Show all steps in the encryption and decryption process; use the fast method for exponentiation. I can do most of the problem except I am having problems proving the fast method, so a detailed explaination on how to do that would be appreciated. Thank you.
- Messages are to be encoded using the RSA method, and the primes chosen are p “ 17 and q “ 19, so that n “ pq “ 323, and e “ 19. Thus, the public key is p323, 19q. (a) Show that the decryption exponent d (your private key) is 91.Let (M, C, e, d, Ke, Ka) denote the RSA cryptosystem with public key (s, n) = (5, 26051) and private key t = 5141. (a) Compute C = e(200, (5, 26051)). (b) Compute M = d(2,5141)Perform calculation of the given problems using the RSA algorithm. i) You intercept the ciphertext C =10 sent to a user whose public key is e=5, n=35. What is the plaintext M?. ii) The public key of a given user is e = 7, N = 33. What is the private key of this user?
- Alice and Bod have decided to use a symmetric encryption algorithm. They have someassumptions about their messages:- Messages only contain capital letters (i.e. A to Z)- The length of their shared key must be greater than or equal to the length of the plaintext- They assign each letter a number as follows: (A,0), (B,1), (C,2), (D,3),…, (Z,25)Their algorithm combines the key and the message using modular addition. The numerical valuesof corresponding message and key letters are added together, modulo 26. For example, if the plaintext is “HELLO” and the key is “SECRET” then the encrypted message is calculated as following:Since the length of the plaintext is 5, we just need the first 5 letters of the key (i.e. “SECRE”), thenfor each letter, we should add corresponding letters in both the plaintext and the key modulo 26.Plaintext: H (7) E (4) L (11) L (11) O (14)Key: S (18) E (4) C (2) R (17) E(4)Cipher: Z (25) I (8) N(13) C(2) S (18) a) Write a program in Python, C/C++ or JavaScript to…Problems are on this Textbook: Stallings, W. (2019). Cryptography and Network Security (8/E). Pearson Publishing. ISBN: 978-0-13576-403-9 7.10 In discussing the CTR mode, it was mentioned that if any plaintext blocks that is encrypted data using a given counter value is known, then the output of the encryption function can be determined easily from the associated cipher text block. SHow the calculation.8. Two of Alice’s public-key components in a simple ElGamal cryptosystem are p = 17 and g =3 . (a) If Alice’s private key is x = 5, what is her public key? (b) What is the ciphertext corresponding to the plaintext p = 4 if the randomly generated number k = 2 is used?
- In the ElGamal cryptosystem, Alice and Bob use p = 17 and α = 3. Bobchooses his secret to be a = 6, so β = 15. Alice sends the ciphertext (r, t) = (7, 6). Determine the plaintext m.We consider ElGamal cryptosystem with public prime p = 53 and generator g = 2. Let h = 16 be Bob’s public-key (his private key x is unknown, h = g^x ). Alice encrypted message m with Bob’s public key and sent the corresponding ciphertext to Bob: (C1 ,C2 ) = (15, 50). You intercepted this ciphertext. Can you break the encryption and recover the original message m? Hint: Both g and h are very small numbers, can you deduce private key x?5. In the RSA system with p = 101, q = 103, and k = 1249, determine the public key(a, n), and encrypt the message M = 2.