HPR0253: Encryption

Three Prime Numbers: 2, 3 and 5.
I will call them A, B and C so A=2, B=3 and C=5
Working through x as the current position, 
and i being the unencrypted password and j being the encrypted as an array,
 so i[x] is the 'current' position we get:

j[x] = ((j[x-1] + i[x] + A) * B ) MOD C

The password example I give is:

Encrypting 123

1 (unencrypted password) + 2 (Prime A) = 3
3 * 3 (Prime B) = 9
9 MOD 5 (Prime C) = 4

4 (previous encrypted) + 2 (current unencrypted) + 2 (Prime A) = 8
8 * 3 (Prime B) = 24
24 MOD 5 (Prime C) = 4

4 (previous encrypted) + 3 (current unencrypted) + 2 (Prime A) = 9
9 * 3 (Prime C) = 27
27 MOD 5 (Prime C) = 2

So the encrypted password is 442

Comments

More Information...