For how many integers n is (n)/(20-n) the square of an integer?

2002 AMC 12B Problem 12 — Solution

2002 AMC 12B Problem 12

Below is the professionally curated solution for Problem 12 of the 2002 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12B solutions, or check the answer key.

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Concepts:perfect square divisibility

Difficulty rating: 1490

12.

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

$1$

$2$

$3$

$4$

$10$

Solution:

Set $\dfrac{n}{20-n}=k^2.$ Solving, $n=\dfrac{20k^2}{k^2+1}.$ Since $k^2$ and $k^2+1$ are coprime, $k^2+1$ must divide $20,$ which happens only for $k=0,1,2,3.$

These give $n=0,$ $10,$ $16,$ $18,$ which is four values.

Thus, the correct answer is D.

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