2002 AMC 12A Problems/Problem 6: Difference between revisions

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

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]]}}
+{{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #6]] and [[2002 AMC 10A Problems|2002 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>?
-<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ }</math>  infinitely many
+<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 ==
-==Solution==
+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==
-For any <math>m</math> we can pick <math>n=1</math>, we get <math>m \cdot 1 \le m + 1</math>,
+https://youtu.be/fo4oSDca-6c
-therefore the answer is <math>\boxed{\text{(E) infinitely many}}</math>.
+~Thesmartgreekmathdude
 ==See Also==
@@ Line 18: / Line 39: @@
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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