Difference between revisions of "2002 AMC 12A Problems/Problem 6"
(New page: ==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>? <math> \mathrm{(A) \ } 4\qquad \math...) |
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+ | {{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #6]] and [[2002 AMC 10A Problems|2002 AMC 10A #4]]}} | ||
+ | |||
==Problem== | ==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>? | 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>? | ||
− | <math> \ | + | <math> \textbf{(A) } 4\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 12\qquad \textbf{(E) } \text{infinitely many} </math> |
+ | |||
+ | ==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{\textbf{(E) } \text{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 infinitely many numbers <math>m</math> that can satisfy the inequality. So the answer is <math>\boxed{\textbf{(E) } \text{infinitely many}}</math>. | ||
+ | |||
+ | ==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 <math>1</math>, therefore the answer is <math>\boxed{\textbf{(E) } \text{infinitely many}}</math>. | ||
+ | |||
+ | ==Video Solution by Daily Dose of Math== | ||
+ | |||
+ | https://youtu.be/fo4oSDca-6c | ||
+ | ~Thesmartgreekmathdude | ||
− | == | + | ==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}} | ||
− | + | [[Category:Introductory Algebra Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 00:44, 19 July 2024
- The following problem is from both the 2002 AMC 12A #6 and 2002 AMC 10A #4, so both problems redirect to this page.
Contents
Problem
For how many positive integers does there exist at least one positive integer n such that
?
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
, therefore the answer is
.
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
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.