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

2002 AMC 12B Problem 12

Attempt Problem 12 of the 2002 AMC 12B below, then check your answer against the professionally curated solution 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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12.

For how many integers

n

\dfrac{n}{20-n}

the square of an integer?

1

2

3

4

10

Set

\dfrac{n}{20-n}=k^2

and solve for

n

in terms of

k

n=\dfrac{20k^2}{k^2+1}

; since

\gcd(k^2,k^2+1)=1,

k^2+1

must divide

20

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.

The signs of

k

do not change

n;

these give

n=0,

10,

16,

and

18,

which is four values.

Thus, the correct answer is D.