Q:

Question about prime factorization of remainders

If $x$ mod $m$ = $y$, and we know the prime factorizations of $x$ and $m$, does this tell us anything about $y$?

A:

We know that $\gcd(x,m)=\gcd(y,m)$, and very little more.

We do know a little bit more.  Let $d=\gcd(x,m)$, $x=dx', m=dm'$, with $\gcd(x',m')=1$.
Since $m|(x-y)$, we also know that $d|(x-y)$ but also $m'|(x-y)$.  Hence if we write $y=dy'$, then we know that $y'\equiv x'\pmod{m'}$.  The primes dividing $y'$ may be entirely different from those primes dividing $x'$, however their product gives the same residue modulo $m'$.

