Homework 16, due Monday, March 7, 2011
 Read pp. 98-99.
 Do Problems 16, 17, p. 108.
 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).