Decrypt the message MAXLJGTEAX which was encrypted using the affine cipher: f(p) (5p+6) mod 26 = Alphabet: A = 0, B = 1, ... , Z = 25 Message:
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- The affine cipher (transformation) C≡3P+24 (mod26) was used to encrypt a message. The resulting ciphertext is RTOLKTOIK. Decrypt this message. A=0,B=1,C=2,...,Y=24,Z=25Decrypt "FYDQAIRR" if it is known to have been encrypted with the affine cipher and c→M and k→C are knownPerform the Encryption and then the Decryption for the Hill Cipher where the Plain text: "POH" and the key is: "GYBNQKURP".
- The ciphertext text “GEZXDS” was encrypted by a Hill cipher with a 2 × 2 matrix. The plaintext is “solved”. Find the encryption matrix M.Decrypt "RXJFKZYH" if it is known to have been encrypted with the affine cipher and c→J and k→N are knownFind the decryption key for the affine cipher with n=26, a=5 and b=8. Encrypt the messagewith the code 15 and 19
- Perform the Encryption using the Hill Cipher where the Plain text: "POH" and the key is: "GYBNQKURP".The affine cipher f(p) = (9p + 10) mod 26 is used to obtain the encrypted message F EMV Z,where p is the two digit representation (00-25) of each character in the original message. What is the original message?Decrypt the word RWLSR if it was encrypted using an alphabetic Caesar shift cipher that starts with shift 9 (mapping A to J), and shifts one additional space after each character is encrypted.
- Encrypt the plaintext “CAT” using the following RSA model if A uses P = 11 and Q = 19 whereas B uses P = 13 and Q = 17. Then Decrypt the cipher to get back the plaintext.Q17. Use the values below to decrypt the numerical ciphertext c = 133 , using the cryptoscheme: Encryption (m + k) mod N = c Decryption (c - k) mod N = m N = 50 k = 42The following cipher text was encrypted using a shift cipher AOPZTLZZHNLPZHMHRL Find the plain text and the key K