1997 AHSME Problems/Problem 13

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Problem

How many two-digit positive integers $N$ have the property that the sum of $N$ and the number obtained by reversing the order of the digits of is a perfect square?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution

Let $N = 10t + u$, where $t$ is the tens digit and $u$ is the units digit.

The condition of the problem is that $10t + u + 10u + t$ is a perfect square.

Simplifying and factoring, we want $11(t+u)$ to be a perfect square.

Thus, $t+u$ must at least be a multiple of $11$, and since $t$ and $u$ are digits, the only multiple of $11$ that works is $11$ itself.

Thus, $(t,u) = (2,9)$ is the first solution, and $(t,u) = (9,2)$ is the last solution. There are $8$ solutions in total, leading to answer $\boxed{E}$.

See also

1997 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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