2002 AMC 10A Problem 4

The question only asks whether some positive integer $n$ works for a given $m$, so try the simplest candidate, $n = 1$. The inequality becomes
$$m \cdot 1 \le m + 1,$$
which holds for every positive integer $m$.

Thus every one of the infinitely many positive integers $m$ has a partner $n$.

The answer is $\boxed{\textbf{(E)}\ \text{infinitely many}}$.

"There exists at least one $n$" is an existence statement, and a single witness settles it. Before doing algebra on an inequality with two unknowns, test the extreme easy values ($n = 1$, $m = 1$); if one of them works uniformly, the problem is over. For comparison, rewriting as $(m-1)(n-1) \le 1$ shows that beyond $n = 1$ only a few pairs work, which is what makes the finite answer choices tempting.

Rearrange: $mn - m - n \le 0$, so $(m-1)(n-1) \le 1$. With $n = 1$ the left side is $0$ for all $m$; with $n = 2$ it forces $m \le 2$; with $n \ge 3$ it forces $m \le 1$ or $m = 2$, $n = 2$. Every $m$ is covered by the $n = 1$ case alone.

• Assuming n must be large or trying to solve the inequality in general instead of noticing that n = 1 always works.
• Counting the pairs $(m, n)$ with $(m-1)(n-1) \le 1$ beyond the $n = 1$ line and reporting a small finite number.