# 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 $m \cdot n \le m + n$? $\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ }$ 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 $\boxed{\text{(E) infinitely many}}$.

### 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 $\boxed{\text{(E) infinitely many}}$.

### 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 $\boxed{\text{(E) infinitely many}}$.

## See Also

 2002 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
 2002 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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