2002 AMC 10B Problem 16

There are infinitely many integers $n$, so first narrow down where $n$ can be. Then check the few candidates.

1. The fraction cannot be negative, because a square is never negative.
- If $n$ is negative, the top is negative and the bottom $20 - n$ is positive, so the fraction is negative.
- If $n$ is more than $20$, the top is positive and the bottom is negative, so again negative.
- If $n = 20$, the bottom is $0$, which is not allowed.

So $n$ is one of $0, 1, 2, \ldots, 19$.
2. How big can the fraction be? As $n$ goes up from $0$ to $19$, the top grows and the bottom shrinks, so the largest value is at $n = 19$: $\frac{19}{1} = 19$. The squares from $0$ to $19$ are $0, 1, 4, 9, 16$.
3. Solve for $n$ for each square. If $\frac{n}{20 - n} = s$, then $n = s(20 - n)$, so $n + sn = 20s$ and
$$n = \frac{20s}{s + 1}.$$

1. The first four are whole numbers; $\frac{320}{17}$ is not. Check one: $n = 18$ gives $\frac{18}{2} = 9 = 3^2$.

So $n = 0, 10, 16, 18$ work: four values.

The answer is $\boxed{\textbf{(D)}\ 4}$.

When a question seems to have infinitely many possibilities, look for limits first. Here the sign of the fraction boxed $n$ into $0$ to $19$, and the size of the fraction limited which squares could appear. After that, a five-column table finished the job. Remember that $0 = 0^2$ counts as a perfect square.

Just test $n = 0$ to $19$. For $n = 1$ to $9$ the top is smaller than the bottom, so the fraction is between $0$ and $1$ and cannot be a whole number. That leaves $n = 0$ and $n = 10$ to $19$:

The squares are at $n = 0, 10, 16, 18$.

Set $\frac{n}{20-n} = k^2$ with $k \ge 0$ an integer. A square is at least $0$, so the fraction cannot be negative. If $n < 0$ the numerator is negative and the denominator positive; if $n > 20$ the reverse; and $n = 20$ is undefined. Hence $0 \le n \le 19$.

On that range the fraction is largest at $n = 19$, where it equals $19$. So $k^2 \le 19$, meaning $k \in \{0, 1, 2, 3, 4\}$.

Solve for $n$: $n = k^2(20 - n)$ gives $n(k^2 + 1) = 20k^2$, so
$$n = \frac{20k^2}{k^2 + 1}.$$
- $k = 0$: $n = 0$.
- $k = 1$: $n = \frac{20}{2} = 10$.
- $k = 2$: $n = \frac{80}{5} = 16$.
- $k = 3$: $n = \frac{180}{10} = 18$.
- $k = 4$: $n = \frac{320}{17}$, not an integer.

Four values of $n$ work: $0, 10, 16, 18$.

The answer is $\boxed{\textbf{(D)}\ 4}$.