Decrypt the message EYDXIOBAYD which was encrypted using the affine cipher: Alphabet: A = 0, B=1,..., 2 25 Message: f(p) = (9p+14) mod 26
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A: The answer to the above-mentioned question is given below with an explanation.
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A: Lets see the solution.
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- 4. Decipher the message ’uxahqomzpuqe’ using a Shift Cipher with k = 12. Show your work.Use symmetric ciphers to encrypt message "“promise" and decrypt message "FOG". The representation of characters in modulo 26 is described as follows: Plaintext - a bcdefg|hijk1 mnopar stuvwx y z Ciphertext A BCD|EFG|HIJKL|M|NO|P|Q|R s T|U|VwxY|Z Value 00 01 02|03 04 05 06|07|08|09|10|11 12|13|14|15 16|17 18|19 20 21 22 23 24 25 The mathematical equations for encryption and decryption can be described as follows: Encryption Ea:i→i+kmod 26 Decryption Da : i→i-k mod 26 i represents the messages (plaintext or cipher), k represents a symmetric key. In this case k=2014- The following message has been encrypted using the Caesar cipher with key 22: ΥΗΚΟΑ PDA ZKΚN. Decrypt it. 15- Prove that for every positive integer n : 1·2· 3 + 2·3· 4+. . +n(n + 1)(n + 2) = n(n + 1)(n + 2)(n + 3)/4
- The message 123 001 324 219 139 has been encrypted with an RSA Public Key Encryption scheme with encryption key (391, 35). Decrypt the message..(c) The following table provides all possible block pairs of plaintext/ciphertext when using some block cipher E with a particular key K. Plaintext Ciphertext Plaintext 0000 1100 1000 1111 1001 0111 1010 1110 1011 1001 0010 1101 0001 0010 0011 0100 0101 0110 0111 1011 1100 1101 1110 1111 Ciphertext 0001 0000 0101 0100 0011 1000 0110 1010 Given that a plaintext P was encrypted using this block cipher operating in a counter mode starting with the counter value of 0, i.e. 0, 1, 2, ..., decrypt the ciphertext C = 010001110111.Encrypt the message "DISCRETE MATHEMATICS" using the shift cipher with k=7
- how to do this; Given an RSA public and private key of (note that not all of the private key is given):N = 3065977e = 425d = 713237 Decrypt the following message that has been encrypted using the public key above: 1559791BLet's encrypt a message using RSA: Choose p = 7 and q = 11, and then select e=13. a.Compute d b.Select plaintext message x=7. Produce the ciphertext y using the fast exponentiation algorithm. c. Decrypt the ciphertext (y) to verify that the initial plaintext (x) is produced. Again, please use the fast exponentiation algorithm.