How to change the modulus of congruence

[arpon]

Suppose we have $a \equiv 0 (mod~ m)$. And, $a \equiv b (mod~ n)$. Is there any way to relate $b$ with $m$ and $n$ ?

 

[Mark44]

Suppose we have $a \equiv 0 (mod~ m)$. And, $a \equiv b (mod~ n)$. Is there any way to relate $b$ with $m$ and $n$ ?

Your first equation is equivalent to a = k₁m, and the second is equivalent to a = k₂n + b.