# 1994 AHSME Problems/Problem 15

## Problem

For how many $n$ in $\{1, 2, 3, ..., 100 \}$ is the tens digit of $n^2$ odd? $\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 50$

## Solution

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 $$10\times 2=\boxed{\textbf{(B) }20.}$$

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