Homework 15, due Friday, March 4, 2011
 Do Problems 1, 7 p. 107.
 Using the encryption method discussed in class I encrypted a plaintext message.
Recall that the encryption works by taking the number x of a letter (as specified below)
and finding the number y of the encrypted letter, using the formula y ≡ xe mod 29.
Here is how we assigned numbers to the letters in class:
a b c d e f g h i j k l m n o p q r s t u v w x y z
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
Here is the encrypted message I got using e = 25 as the encryption exponent:
meb fxvb, oujb czyub buaxeyu nxglha ozglxo mahjkf? zb zf buh hjfb jgl reczhb zf buh feg.
The decryption exponent d turns out to be 9. I.e., you can find x given y using the formula
x ≡ yd mod 29 with d = 9.
For example, m is letter number 14, so to find what letter it was originally, we compute 149 (mod 29), using the web form below.
It tells us that 149 ≡ 3 (mod 29), and 3 is the number of the letter b. Thus the m was originally a b.
Use the web form below to decrypt enough of this ciphertext to identify the source.
Just plug in the variables m, n and a, and you will get back
the value of ma (mod n).