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 first digit of $100a^2$ it the hundreds place digit which means that $20ab+b^2$ is the tens and ones digit together. We note that $20ab$ will always be even. So in order for $20ab+b^2$ to have an odds tens digit, then $b^2$ must carry and have an odd tens digit. We note that $b$ can only equal either $4$ or $6$ for this to happen. 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

1994 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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1994 AHSME Problems/Problem 15

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