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.)

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(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 = 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 = 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

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.)
Transcribed Image Text: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.)
(1415) ¹¹ = (1415)8+2+1 = (1415)8 (1415)2 (1415) mod 2911.
In detail, we obtain
Hence we have
1415² = 2002225 = 2368 mod 2911
14154 = 2368² = 5607424 = 838 mod 2911
14158 = 838² = 702244 = 693 mod 2911.
(1415) ¹¹ = (693) (2368) (1415) mod 2911
= (2131) (1415) mod 2911
= 2480 mod 2911.
Thus the message "NO" is encrypted as 2480 and that is what is sent to me.
This took a little time with a calculator, but of course a computer would do
it very quickly. <
Now for the decryption. Since e was chosen to be relatively prime to
p - 1 and to q - 1, it is relatively prime to t, their least common multiple.
Thus, by Proposition (2.6), è is a unit in Zt. The key to decryption is to
find the multiplicative inverse of è in this ring.
Transcribed Image Text:(1415) ¹¹ = (1415)8+2+1 = (1415)8 (1415)2 (1415) mod 2911. In detail, we obtain Hence we have 1415² = 2002225 = 2368 mod 2911 14154 = 2368² = 5607424 = 838 mod 2911 14158 = 838² = 702244 = 693 mod 2911. (1415) ¹¹ = (693) (2368) (1415) mod 2911 = (2131) (1415) mod 2911 = 2480 mod 2911. Thus the message "NO" is encrypted as 2480 and that is what is sent to me. This took a little time with a calculator, but of course a computer would do it very quickly. < Now for the decryption. Since e was chosen to be relatively prime to p - 1 and to q - 1, it is relatively prime to t, their least common multiple. Thus, by Proposition (2.6), è is a unit in Zt. The key to decryption is to find the multiplicative inverse of è in this ring.
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