Homework 15, due Friday, March 4, 2011

[1] Do Problems 1, 7 p. 107.

[2] 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).

Enter m:

Enter a:

Enter n: