Using the plaintext "BLUE SKY" demonstrate the use of 5 as the multiplicative key to generate the substitution ciphertext.

Use the modified Caesar cipher described in the assignment instructions to decrypt the following:Ciphertext:  UCTOTBWVUCKey: 5

Give me answer please.

(1) Find a quote that you like. Make sure that the total number of characters is at least 50 and no more than 100.  (2) Then pick a number that is between 4 and 24 as your additive key.  (3) Create a table of correspondence of 26 alphabets in plaintext and the ciphertext.  (4) Use the table of correspondence to encode the quote you found in Part (1). Post the encrypted message:For example, here is a quote from John DeweyFailure is instructive. The person who really thinks learns quite as much from his failures as from his successes.We will use additive key 17. The table of correspondence between plaintext and ciphertext is as follows.(attached)      Thus the encrypted message isWRZ CLI VZJ ZEJ KIL TKZ MVK YVG VIJ FEN YFI VRC CPK YZE BJC VRI EJHLZK VRJ DLT YWI FDY ZJW RZC LIV JRJ WIF DYZ JJL TTV JJV J(5) Pick an encrypted message from your classmate and try to decrypt it. Remember to use the frequency table because you will not have access to the additive key. Make sure to include all…

45. Encrypt the message “success is not final, failure is not fatal” using a columnar transposition cipher for the given keyword “algebra” (1 6 5 4 3 7 2)..

If the shared key is "SECURITY" and the result of DES round 15 is "IT? 7b/," what is the ciphertext?

Decode the number using the given private key. Decode the number M = 15 using the private key d = 53 and n = 77. Assume you are decoding the number using the RSA cryptosystem. Note: You can use the Modular Exponentiation calculator to help with the calculation. a 63 (mod 77), decoded number = 63 b 64 (mod 77), decoded number = 64 C 65 (mod 77), decoded number = 65 d. 66 (mod 77), decoded number = 66

Could you devise a key that when it is used on the ciphertext you produced in 1.a, will provide plaintext relish instead of the strike? the photo below is question 1a and please trying to be as detailed as possible when answering the question, justifying well all your steps that you did.

Following the example, please break this cipher:   c=6642328489179330732282037747645 n=17058317327334907783258193953123 Please use screenshots to demonstrate the intermediate steps and results.   [screenshot]

9. Use the additive cipher with k = 5 to encrypt the plaintext 'MTCS'.