For how many $n$ in $\{1, 2, 3, ..., 100 \}$ is the tens digit of $n^2$ odd?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 50$
Let $n=10a+b$. So $n^2=(10a+b)^2=100a^2+20ab+b^2$. The term $100a^2$ only contributes digits starting at the hundreds place, so this does not affect whether the tens digit is odd.  The term $20ab$ only contributes digits starting at the tens place, and the tens digit contributed will be the ones digit of $2ab$ which is even.  So we see this term also does not affect whether the tens digit is odd.  This means only $b^2$ can affect whether the tens digit is odd.  We can quickly check $1^2=1, \dots, 9^2=81$ and discover only for $b=4$ or $b=6$ does $b^2$ have an odd tens digit. The total number of positive integers less than or equal to $100$ that have $4$ or $6$ as the units digit is \[10\times 2=\boxed{\textbf{(B) }20.}\]
--Solution by [TheMaskedMagician](https://artofproblemsolving.comhttp://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685)