Solving $217 x \equiv 1 \quad \text{(mod 221)}$ I am given the problem:
Find an integer $x$ between $0$ and $221$ such that
$$217 x \equiv 1 \quad \text{(mod 221)}$$
How do I solve this? Unfortunately I am lost.

In this special case, you can multiply the congruence by $-1$ and you'll get
$$4x\equiv 220 \pmod{221}.$$
(Just notice that $-217 \equiv 4 \pmod{221}$ and $-1\equiv220\pmod{221}$.)
This implies that $x\equiv 55 \pmod{221}$ is a solution. (And since $\gcd(4,221)=1$, there is only one solution modulo $221$.)

In general, for questions of this type you can use extended Euclidean algorithm see Wikipedia.
You can find some examples at this site, e.g. here.
