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University of South Carolina High School Math Contest/1993 Exam/Problem 9

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Problem

Suppose that $x$ and $y$ are integers such that $y > x > 1$ and $y^2 - x^2 = 187$ . Then one possible value of $xy$ is

$\mathrm{(A) \ }30 \qquad \mathrm{(B) \ }36 \qquad \mathrm{(C) \ }40 \qquad \mathrm{(D) \ }42 \qquad \mathrm{(E) \ }54$

Solution

We have $(y+x)(y-x)=187$ . Now, $187=11 \cdot 17$ , so we have as one possibility $y+x=17$ and $y-x=11$ . Thus, $x=3, y=14$ is a possible solution and the answer is $42$ .

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Category: Introductory Algebra Problems

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