Difference between revisions of "2002 AMC 12A Problems/Problem 6"
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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{\text{(E) 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 1, therefore the answer is <math>\boxed{\text{(E) infinitely many}}</math>. | ||
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==See Also== | ==See Also== | ||
Revision as of 21:28, 1 January 2019
- 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 
does there exist at least one positive integer n such that 
?
infinitely many
Solution
Solution 1
For any we can pick 
, we get 
,
therefore the answer is 
.
Solution 2
Another solution, slightly similar to this first one would be using Simon's Favorite Factoring Trick.
Let , then
This means that there are infinitely many numbers that can satisfy the inequality. So the answer is .
Solution 3
If we subtract 
from both sides of the equation, we get 
. Factor the left side to get 
. Divide both sides by 
and we get 
. The fraction 
if 
. There is an infinite amount of integers greater than 1, therefore the answer is .
See Also
| 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. 
