It will also be useful to switch back and forth between congruences and regular equations. The above fact helps with this. We know that \(a \equiv b \pmod{n}\) if and only if \(n \mid a-b\text{,}\) if and only if \(a-b = kn\) for some integer \(k\text{.}\) Rearranging that equation, we get \(a = b + kn\text{.}\) In other words, if \(a\) and \(b\) are congruent modulo \(n\text{,}\) then \(a\) is \(b\) more than some multiple of \(n\text{.}\) This conforms with our earlier observation that all the numbers in a particular remainder class are the same amount larger than the multiples of \(n\text{.}\)
in-context