# 2002 AMC 12A Problems/Problem 6

The following problem is from both the 2002 AMC 12A #6 and 2002 AMC 10A #4, so both problems redirect to this page.

## Problem

For how many positive integers $m$ does there exist at least one positive integer n such that $m \cdot n \le m + n$? $\textbf{(A) } 4\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 12\qquad \textbf{(E) } \text{infinitely many}$

## Solution

### Solution 1

For any $m$ we can pick $n=1$, we get $m \cdot 1 \le m + 1$, therefore the answer is $\boxed{\textbf{(E) } \text{infinitely many}}$.

### Solution 2

Another solution, slightly similar to this first one would be using Simon's Favorite Factoring Trick. $(m-1)(n-1) \leq 1$

Let $n=1$, then $0 \leq 1$

This means that there are infinitely many numbers $m$ that can satisfy the inequality. So the answer is $\boxed{\textbf{(E) } \text{infinitely many}}$.

### Solution 3

If we subtract $n$ from both sides of the equation, we get $m \cdot n - n \le m$. Factor the left side to get $(m - 1)(n) \le m$. Divide both sides by $(m-1)$ and we get $n \le \frac {m}{m-1}$. The fraction $\frac {m}{m-1} > 1$ if $m > 1$. There is an infinite amount of integers greater than $1$, therefore the answer is $\boxed{\textbf{(E) } \text{infinitely many}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 