5.7. Suppose n = 2911 and e = 11. Encrypt the following messages as in Example (5.3). a) "OK" b) "HELP" (Break this up into two blocks.)
(5.3) EXAMPLE. Of course, in the examples we will present in this section to illustrate these ideas, we will not use a 200-digit integer for n. I announce my integers to be n = 2911 and e=11. (The reader can obtain the prime factorization of n without too much difficulty; we will give it below.) Suppose someone wants to send me the message "NO." This word is first converted into the integer M = 1415 by our simple system above. The sender then computes (1415)^11 mod 2911. This may seem like a very difficult computation itself, but it is not too bad if we proceed as follows. We compute (1415)^2 mod 2911, then we square this to obtain (1415)^4 mod 2911, then square again to obtain (1415)^8 mod 2911. Now we use the fact that 11= 8+2+1 and so [See image] (R. F. Lax, Modern Algebra and Discrete Structures, HarperCollins, 1991. Accessed 9 March, 2023)
PLEASE HELP WITH QUESTION 5.7



Trending now
This is a popular solution!
Step by step
Solved in 3 steps









