Find the least nonnegative residue of 2(100!) modulo 103
This is what I have so far but dont know where to go form here:
2(100!)x≡1 mod 103
2(100!)x≡104 mod 103
100!x≡52 mod 103
Keep in mind that we want to find x in
We know that , that is to say
So if I can reduce , it would be nice !
101 is 103-2, so writing it this way will make appear 103.
Hence, if we go back to the boxed expression, we'll get :
But we know that (1)