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

Revision as of 20:05, 11 July 2009

The following problem is from both the 2002 AMC 12A #6 and 2002 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 $\boxed{\text{(E) infinitely many}}$ .

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 infinately many numbers $m$ that can satisfy the inequality. So 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 7: / Line 7: @@
-==Solution==
+==Solution 1==
 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>.
+==Solution 2==
+Another solution, slightly similar to this first one would be using Simon's Favorite Factoring Trick.
+<math>(m-1)(n-1) \leq 1</math>
+Let <math>n=1</math>, then
+<math>0 \leq 1</math>
+This means that there are infinately many numbers <math>m</math> that can satisfy the inequality. So the answer is <math>\boxed{\text{(E) infinitely many}}</math>.
 ==See Also==