jmc

We use the property $a\equiv b(\mathrm{mod}m)$ implies $ac\equiv bc(\mathrm{mod}m)$.

Since all numbers with the units digit of $2$ has a remainder of $2$ when divided by $5$ and we have $10$ numbers, $$2\times 12\times 22\times 32\times \dots \times 72\times 82\times 92\equiv 2^{10}\equiv 1024\equiv \boxed{4}(\mathrm{mod}5).$$