How to solve $\{ x \in \mathbb N_0 | x \cdot416 \equiv 1696 \pmod {2208}\}$?
Since $416$ is not invertible in $2208$ I have cancelled everything by $\gcd(416, 2208, 1696)= 32$.
That gives:
$\{ x \in \mathbb N_0 \mid x \cdot13 \equiv53 \pmod {69} \}$.
So I have calculated $13^{-1} $ mod $69 = 16$.
That is the wrong result. The correct answer would be $20$.
Question: What is the way to solve that task?