On the other hand, if we assume first that \(n \mid a-b\text{,}\) so \(a-b = kn\text{,}\) then consider what happens if we divide each term by \(n\text{.}\) Dividing \(a\) by \(n\) will leave some remainder, as will dividing \(b\) by \(n\text{.}\) However, dividing \(kn\) by \(n\) will leave 0 remainder. So the remainders on the left-hand side must cancel out. That is, the remainders must be the same.
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