@Jhevon : this is number theory
We're working on integers
Since 11 and 1476 are coprime, there exists u and v that verify this equality (Bézout's identity)
Now use the Euclidian algorithm, which lets us say that at a moment, there will be 1 as the reminder :
And this last expression gives you u and v
If you're looking for the general solution, tell us !
I think that is what i need....I needed to solve for u, I still can't do the problem because now the equation looks like this:
a^u = = b (mod m) where a is my list to decode and u=671 m=1557
I'm lost when it comes to raising somthing to the 671 power....and my formula for mods it would take me weeks....So I just can't figure this one out : (
The problem is that
If p is prime,
But I don't think you have to do it this way... You can use the fact that if a is coprime (I don't remember what happens if it is not) with 1557, the order of a in , that is to say the least integer m such that , divides 1032.
How long is your list ?