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

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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>?
  
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<math> \textbf{(A) \ } 4\qquad \textbf{(B) \ } 6\qquad \textbf{(C) \ } 9\qquad \textbf{(D) \ } 12\qquad \textbf{(E) \ }</math> infinitely many
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<math> \textbf{(A) } 4\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 12\qquad \textbf{(E) } \text{infinitely many} </math>
  
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==Solution==
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==Solution 1 ==
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===Solution 1===
 
  
 
For any <math>m</math> we can pick <math>n=1</math>, we get <math>m \cdot 1 \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>,
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therefore the answer is <math>\boxed{\text{(E) infinitely many}}</math>.
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therefore the answer is <math>\boxed{\textbf{(E) } \text{infinitely many}}</math>.
  
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==Solution 2 ==
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===Solution 2===
 
  
 
Another solution, slightly similar to this first one would be using [[Simon's Favorite Factoring Trick]].
 
Another solution, slightly similar to this first one would be using [[Simon's Favorite Factoring Trick]].
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<math>0 \leq 1</math>
 
<math>0 \leq 1</math>
  
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This means that there are infinitely many numbers <math>m</math> that can satisfy the inequality. So the answer is <math>\boxed{\text{(E) infinitely many}}</math>.
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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>.
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==Solution 3 ==
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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 <math>1</math>, therefore the answer is <math>\boxed{\textbf{(E) } \text{infinitely many}}</math>.
  
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==Video Solution by Daily Dose of Math==
  
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===Solution 3===
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https://youtu.be/fo4oSDca-6c
  
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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>.
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~Thesmartgreekmathdude
  
 
==See Also==
 
==See Also==

Latest revision as of 23:44, 18 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.

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

Video Solution by Daily Dose of Math

https://youtu.be/fo4oSDca-6c

~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. AMC logo.png

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