Solvability of $ax\equiv b\pmod n$

Necessity. If $ax\equiv b\pmod n$, then $ax-b=kn$, so $d\mid ax-kn$, hence $d\mid b$.

Sufficiency. Suppose $d\mid b$, say $b=db'$. Write $a=da'$, $n=dn'$. Then $a'x\equiv b'\pmod{n'}$ with $\gcd(a',n')=1$. Since $a'$ is invertible mod $n'$, there is a unique solution $x\equiv x_0\pmod{n'}$.

Lifting. The solutions mod $n$ are $x_0, x_0+n', x_0+2n',\ldots,x_0+(d-1)n'$: exactly $d$ solutions.

Conclusion. $ax\equiv b\pmod n$ has $d=\gcd(a,n)$ solutions mod $n$ iff $d\mid b$.