Homework 16, due Monday, March 7, 2011

[1] Read pp. 98-99.

[2] Do Problems 16, 17, p. 108.

[3] Using the encryption method discussed in class I encrypted a plaintext message, but this time I used e = 11 as the encryption exponent, so now 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 = 11 as the encryption exponent:

bxxl tabuy gxxt

To decrypt this message you must find the decryption exponent d. To find it you must solve the equation d*e ≡ 1 (mod 28). Since e = 11, this is: 11d ≡ 1 (mod 28). I.e, you must find a multiple of 11 that leaves a remainder of 1 when divided by 28. When you have d, you can then decrypt using
x ≡ yd mod 29.

You can use the web form below to do the arithmetic. So the problem is to find d and determine what the encrypted message says.

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: