working code examples: A47X3,R28X2,X21X9, T49X1

The first Char is a random letter (it can not be "O" so people wont mistake it for 0)

The second is a number between 1 and 4

the third is a number between 1 and 9

The fourth is always X

The fifth is a number between 1 and 9

so 25x4x9x9 = 8100 WOULD be my answer but there is one more rule to this code that complicates it.

The second Char + the First third char + the fifth char must equal either 12 or 14. I do not know how to calculate the number of combinations taking this into consideration.

Mods, please move my post if there is a better area for it.

This post has been edited by **xilith117**: 10 February 2014 - 10:48 PM