Let $m$ successive integers $a,a+1,a+2,\cdots,a+m-1$ and another interger $A$ be given,then one,and only one,of these integers will be congruent to to $A$ relative to $m$.
Proof:Existence:We just need to prove that there exists $k\in\mathbf{Z}$,such that $A+mk\in \{a,a+1,a+2,\cdots,a+m-1\}$.If there is no such $k$,then
\begin{equation}
m\not|(a-A),m\not|(a+1-A),\cdots,m\not|(a+m-1-A)
\end{equation}
This is absurd(Why?)
Uniqueness:Trival(Why?).