Complete Residue System Modulo m has m Elements

Theorem
Let $m \in \Z_{\ne 0}$ be a non-zero integer.

Let $S := \set {r_1, r_2, \dotsb, r_s}$ be a complete residue system modulo $m$.

Then $s = m$.

Proof
Let:
 * $t_0 = 0, t_1 = 1, \dots, t_{m - 1} = m - 1$

Let $n \in \Z$.

Then from the Division Theorem there exist unique integers $q$ and $u$ such that:
 * $n = m q + u$

such that $0 \le u < m$.

That is:
 * $n \equiv u \pmod m$

and $u$ is one of $t_0, t_1, \ldots, t_{m - 1}$.

Also, since $\size {t_i - t_j} < m$, no two elements of $\set {t_0, t_1, \ldots, t_{m - 1} }$ are congruent.

Thus $\set {t_0, t_1, \ldots, t_{m - 1} }$ is a complete residue system modulo $m$.

So each $r_i$ is congruent to exactly one element of $\set {t_0, t_1, \ldots, t_{m - 1} }$.

So:
 * $s \le m$

And since $\set {r_1, r_2, \dotsb, r_s}$ forms a complete residue system modulo $m$, every element of $\set {t_0, t_1, \ldots, t_{m - 1} }$ is congruent to exactly one element of $S$.

So:
 * $m \le s$

Hence the result.