jmc

Let $d$ be the last digit of a number $n$. Then $n^2\equiv d^2(\mathrm{mod}10)$, so the units digit of $n^2$ is the same as the units digit of $d^2$. Checking all the digits from 0 to 9, we find that the possible units digits of $d^2$ are 0, 1, 4, 5, 6, and 9, for a total of $\boxed{6}$.