2002 AMC 10B Problem 16

Attempt Problem 16 of the 2002 AMC 10B 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 10B solutions, or check the answer key.

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16.

For how many integers

n

is

\dfrac{n}{20 - n}

the square of an integer?

1

2

3

4

10

Answer: D

Concepts:Diophantine Equation divisibility perfect square

Difficulty rating: 1580

Small hint:

Set

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

and solve for

n

in terms of

k

Big hint:

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

; since

k^2

and

k^2 + 1

are coprime,

20

is divisible by

k^2 + 1

Solution:

Suppose

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

for some integer

k \ge 0.

Solving,

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

Since

k^2

and

k^2 + 1

share no common factor,

k^2 + 1

must divide

20.

This happens only for

k = 0,

1,

2,

3,

giving

k^2 + 1 = 1,

2,

5,

10.

The corresponding values

n = 0,

10,

16,

18

are all integers, so there are

4

such

n.

Thus, the correct answer is D.