There's a proof in a section in my book on modular arithmetic I don't really understand. On linear congruences:
$ax \equiv b \text{ (mod $m$)}$
Suppose that $\gcd({a,m}) = d > 1$ and $b$ is divisible by $d$. We can divide both sides of the linear congruence by $d$:
$\frac{a}{d}x \equiv \frac{b}{d}$ (mod $\frac{m}{d}$), gcd($\frac{a}{d}$, $\frac{m}{d}$) $= 1$.
This last linear congruence has exactly one solution $r \in$ $\mathbb{N}[0, \frac{m}{d}-1]$. Therefore the solutions to $ax \equiv$ $b$ (mod $m$) are $r + t\frac{m}{d}$ with $t \in \mathbb{N}[0,d-1]$.
I understand everything except the last sentence. I have an intuition on why this might be true, but how can it be proven?