Using RSA, calculate the ciphertext where: a. Plaintext = 20, e =21, and n =300 You should show all details including the fast exponentiation calculation

A range check can be used to make sure that values fall between a specified minimum and maximum value. O True False

7. Select primes p=1009, q=1013, And find the encrypted and decrypted key using RSA algorithm.

(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…