2002 AMC 10B Problem 16

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 $\boxed{\textbf{(D)}\ 4}$.

Bounding first shrinks an infinite search to a handful of cases: the sign of the fraction limits $n$, and the size of the fraction limits $k$. Once the search space is tiny, solving for $n$ in terms of $k$ and checking integrality is faster than any cleverness. Remember that $0$ counts as a perfect square.

Divisibility: from $n = \frac{20k^2}{k^2+1}$ and $\gcd(k^2, k^2+1) = 1$, the integer $k^2 + 1$ must divide $20$. The divisors $1, 2, 5, 10$ give $k^2 = 0, 1, 4, 9$ (while $4$ and $20$ give the non-squares $3$ and $19$), so again $k = 0, 1, 2, 3$.

• Overlooking n = 0 (the square 0^2) or assuming a 'square of an integer' must be positive, which drops the count to 3.
• Counting $k$ and $-k$ as separate solutions (they give the same $n$), or forgetting that $n = 20$ makes the expression undefined.