Difference between revisions of "2002 AMC 12A Problems/Problem 6"

Revision as of 12:35, 8 November 2021

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

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 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 by Problem 3	Followed by Problem 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

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

@@ Line 26: / Line 26: @@
 ===Solution 3===
-If we subtract <math>n</math> from both sides of the equation, we get <math>m \cdot n - n \le m</math>. Factor the left side to get <math>(m - 1)(n) \le m</math>. Divide both sides by <math>(m-1)</math> and we get <math> n \le \frac {m}{m-1}</math>. The fraction <math>\frac {m}{m-1} > 1</math> if <math>m > 1</math>. There is an infinite amount of integers greater than 1, therefore the answer is <math>\boxed{\textbf{(E) } \text{infinitely many}}</math>.
+If we subtract <math>n</math> from both sides of the equation, we get <math>m \cdot n - n \le m</math>. Factor the left side to get <math>(m - 1)(n) \le m</math>. Divide both sides by <math>(m-1)</math> and we get <math> n \le \frac {m}{m-1}</math>. The fraction <math>\frac {m}{m-1} > 1</math> if <math>m > 1</math>. There is an infinite amount of integers greater than <math>1</math>, therefore the answer is <math>\boxed{\textbf{(E) } \text{infinitely many}}</math>.
 ==See Also==

Page

Toolbox

Search