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

Revision as of 07:56, 18 February 2009

The following problem is from both the 2009 AMC 12A #6 and 2009 AMC 10A #4, so both problems redirect to this page.

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

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

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2002 AMC 12A Problems|2009 AMC 12A #6]] and [[2002 AMC 10A Problems|2009 AMC 10A #4]]}}
 ==Problem==
 For how many positive integers <math>m</math> does there exist at least one positive integer n such that <math>m \cdot n \le m + n</math>?
@@ Line 6: / Line 8: @@
 ==Solution==
-<math>m \cdot 1 \le m + 1</math> for <math>n=1</math>
-<math>m \le m + 1</math>
+For any <math>m</math> we can pick <math>n=1</math>, we get <math>m \cdot 1 \le m + 1</math>,
+therefore the answer is <math>\boxed{\text{(E) infinitely many}}</math>.
+==See Also==
+{{AMC12 box|year=2002|ab=A|num-b=5|num-a=7}}
+{{AMC10 box|year=2002|ab=A|num-b=3|num-a=5}}
-<math> \Rightarrow \mathrm{(E) \ }</math> infinitely many
+[[Category:Introductory Algebra Problems]]