Let's say I do this: B8 Mod 19 and get 9..... What is the fastest possible way to calculate that B8 back ......having only the outcome of B8 Mod 19 (9) and the number B8 was moded with?
My solution is up there however, its a bit costly in cycles

I'm not sure that I understood you correctly...
Do you mean that you have numbers 9 and 19 and you want to find X, where X mod 19 = 9?
If so then there is not only B8 which will give you 9 when moded with 19. What operand sizes are used? One byte only?

Simple answer: you can't!

You only know B8%19 and 19, but you don't know B8 neither 7. You have to know at least one of them.

When you've got 9 and 19 then the B8 can be anything from 9, 28, ... to 19*n+9.

For the solution - if you have the neccessary variables - yours is correct and not very costy in cycles.

Ok in decimal not in hex my solution works.

a mod b = c

a = [(b * (a/b)] + (c) " without the remainder of a/b

Ex : 99 mod 19 = 4


99 = (19*(a/b)) +4

Hwoever I dont have a

Im using an encryption for a program based on modolus .. And I need a way to find a..... This is the only way I have .... But with a... But in the program I won't be able to have a

It will keep going like this.....

One of the properties of congruences is symmetry, that's:
if a = b (mod m), then b = a (mod m).
Things are very simple when m = 2. It's just XOR, you know.
But if modulo != 2 than you cannot do it in just a few cycles. The simplest solution that comes to my mind right now is:
1. convert numbers to base m (19 in your example)
2. apply XOR equivalent to them.
3. conevrt back to base 2 (or 16 - it doesn't matter).

or you can work with "base m" numbers all the time. some math functions should be implemented then. But I cannot give the optimal solution at this time. The proplem is interesting and i'll think of it for sure... when I have time.

visit http://mathworld.wolfram.com/.

BTW, you'll have more than one solution of this problem. Maybe selecting the minimum one (or, for example, the second from the mininum) for the chosen operand size is the way to avoid errors when decipher.

I was trying to solve the problem, but was confused a bit. But now I think I've got the idea of what you wanna do. Is it a multiplicative cipher you're trying to implement? Please, provide the community with your algorithm

If it is so, then the simplest form of a multiplicative cipher is:
encipher: ciphertext=msg*key (mod m)
decipher: msg=ciphertext*(1/key) (mod m)

to find 1/key (mod m) you should use Eucledian algo. I can provide you with formulas to implement it... if you need.

zhak, we will be very appreciate for this formulas, i am also interested in this problem

(input mod key)-0C4h)

This is what Im using

I really can't solve it sigh...

I believe it's only possible by a bruteforce of such

Vasilev Vjacheslav, here's how to find 1/a (mod b)

So do you have a solution for my problem?