2002 AMC 10A Problem 4

Read the question carefully: for each $m$, we only need one positive integer $n$ that works. So try the easiest possible $n$.

1. Try an example. Take $m = 5$ and $n = 1$. Then $m \cdot n = 5$ and $m + n = 6$. Since $5 \le 6$, the rule holds, so $m = 5$ counts.

2. Try it for any $m$. With $n = 1$, the left side is $m \cdot 1 = m$ and the right side is $m + 1$. A number is always less than that number plus one, so
$$m \le m + 1$$
is true for every positive integer $m$.

1. So every positive integer $m$ has a partner, namely $n = 1$. There are infinitely many positive integers.

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

"There exists at least one $n$" means a single example is enough. Before doing algebra with two unknowns, test the simplest values like $1$; if one of them works for every case, you are done. The small numbers in the answer choices are there to tempt students who count all working pairs instead.

Move everything to one side: $mn - m - n \le 0$. Adding $1$ to both sides lets the left side factor:
$$mn - m - n + 1 = (m - 1)(n - 1), \quad\text{so}\quad (m-1)(n-1) \le 1.$$
When $n = 1$ the left side is $(m-1) \times 0 = 0$, which is at most $1$ for every $m$. So again every $m$ works. (Other values of $n$ work only for a few small $m$, for example $m = n = 2$.)