Difference between revisions of "2002 AMC 12B Problems/Problem 12"

Revision as of 16:30, 18 January 2008

Problem

For how many integers $n$ is $\dfrac n{20-n}$ the square of an integer?

$\mathrm{(A)}\ 1 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 3 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 10$

Solution

Let $x^2 = \frac{n}{20-n}$ , with $x \ge 0$ (note that the solutions $x < 0$ do not give any additional solutions for $n$ ). Then rewriting, $n = \frac{20x^2}{x^2 + 1}$ . Since $\text{gcd}(x^2, x^2 + 1) = 1$ , it follows that $x^2 + 1$ divides $20$ . Listing the factors of $20$ , we find that $x = 0, 1, 2 , 3$ are the only $4 \Rightarrow \mathrm{(D)}$ solutions (respectively yielding $n = 0, 10, 16, 18$ ).

2002 AMC 12B (Problems • Answer Key • Resources)
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Difference between revisions of "2002 AMC 12B Problems/Problem 12"

Revision as of 16:30, 18 January 2008

Problem

Solution

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