107,105 point musings

Relevant wiki: Perfect Squares, Cubes, and Powers

From $101\:101 \leq n^2 \leq 109\:909$ it follows by taking square roots that $318 \leq n \leq 331.$ We can rule out $320$ and $330$ immediately because their squares would end in zero; and $325$ because its square would end in $25$ .

For the remaining 11 numbers, it helps to know that the last two digits of the square is equal to the last two digits of the square of the last two digits; e.g. $318^2 \equiv 18^2 \equiv 24$ mod 100, etc.

As it turns out, none of the squares of these numbers has a zero at the tens position. Therefore the answer is $\boxed{\text{zero, zilch, zip, naught, nada}}$ .