2002 AMC 10B Problem 16

P16-20~4 minPrint

Diophantine Equations Number Properties Bound the quantity above/below or estimate to pin it down Organized listing / direct enumeration

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

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Key insight

A square is never negative, so n is from 0 to 19 and the fraction is at most 19. Only the squares 0, 1, 4, 9 give whole-number n.

Solution

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
$n = \frac{20s}{s + 1}.$

square $s$	$0$	$1$	$4$	$9$	$16$
$n = \frac{20s}{s+1}$	$0$	$10$	$16$	$18$	$\frac{320}{17}$

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

Why this works

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.

Alternative approach

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 $19$ :

$n$	$0$	$10$	$11$	$12$	$13$	$14$	$15$	$16$	$17$	$18$	$19$
$\frac{n}{20-n}$	$0$	$1$	$\frac{11}{9}$	$\frac32$	$\frac{13}{7}$	$\frac73$	$3$	$4$	$\frac{17}{3}$	$9$	$19$

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

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Number Theory, Chapter 7: Algebra With Integers
Introduction to Number Theory, Chapter 1: Integers: The Basics
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Algebra, Chapter 11: Special Factorizations (§11.5)