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2008 AMC 12A Problems/Problem 5: Difference between revisions

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New page: ==Problem== Suppose that <math>\frac {2x}{3} - \frac {x}{6}</math> is an integer. Which of the following statements must be true about <math>x</math>? <math>\textbf{(A)}\ \text{It is ne...
 
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{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #5]] and [[2008 AMC 10A Problems/Problem 9|2008 AMC 10A #9]]}}
==Problem==
==Problem==
Suppose that
Suppose that
 
<cmath>\frac{2x}{3}-\frac{x}{6}</cmath>
<math>\frac {2x}{3} - \frac {x}{6}</math>
 
is an integer. Which of the following statements must be true about <math>x</math>?
is an integer. Which of the following statements must be true about <math>x</math>?


<math>\textbf{(A)}\ \text{It is negative.} \qquad \textbf{(B)}\ \text{It is even, but not necessarily a multiple of }3\text{.} \\
<math>\mathrm{(A)}\ \text{It is negative.}\\\mathrm{(B)}\ \text{It is even, but not necessarily a multiple of 3.}\\\mathrm{(C)}\ \text{It is a multiple of 3, but not necessarily even.}\\\mathrm{(D)}\ \text{It is a multiple of 6, but not necessarily a multiple of 12.}\\\mathrm{(E)}\ \text{It is a multiple of 12.}</math>
\textbf{(C)}\ \text{It is a multiple of }3\text{, but not necessarily even.} \\
\textbf{(D)}\ \text{It is a multiple of }6\text{, but not necessarily a multiple of }12\text{.} \\
\textbf{(E)}\ \text{It is a multiple of }12\text{.}</math>


==Solution==  
==Solution==
Since <math>\frac {2x}{3} - \frac {x}{6} = \frac{4x}{6}-\frac{x}{6}=\frac{3x}{6}=\frac{x}{2}</math> is an integer, <math>x</math> must be even <math>\Rightarrow B</math>.  
<cmath>\frac{2x}{3}-\frac{x}{6}\quad\Longrightarrow\quad\frac{4x}{6}-\frac{x}{6}\quad\Longrightarrow\quad\frac{3x}{6}\quad\Longrightarrow\quad\frac{x}{2}</cmath>
For <math>\frac{x}{2}</math> to be an integer, <math>x</math> must be even, but not necessarily divisible by <math>3</math>. Thus, the answer is <math>\mathrm{(B)}</math>.


==See Also==  
==See also==
{{AMC12 box|year=2008|num-b=4|num-a=6|ab=A}}
{{AMC12 box|year=2008|ab=A|num-b=4|num-a=6}}
{{AMC10 box|year=2008|ab=A|num-b=8|num-a=10}}
{{MAA Notice}}

Latest revision as of 02:17, 5 August 2021

The following problem is from both the 2008 AMC 12A #5 and 2008 AMC 10A #9, so both problems redirect to this page.

Problem

Suppose that \[\frac{2x}{3}-\frac{x}{6}\] is an integer. Which of the following statements must be true about $x$?

$\mathrm{(A)}\ \text{It is negative.}\\\mathrm{(B)}\ \text{It is even, but not necessarily a multiple of 3.}\\\mathrm{(C)}\ \text{It is a multiple of 3, but not necessarily even.}\\\mathrm{(D)}\ \text{It is a multiple of 6, but not necessarily a multiple of 12.}\\\mathrm{(E)}\ \text{It is a multiple of 12.}$

Solution

\[\frac{2x}{3}-\frac{x}{6}\quad\Longrightarrow\quad\frac{4x}{6}-\frac{x}{6}\quad\Longrightarrow\quad\frac{3x}{6}\quad\Longrightarrow\quad\frac{x}{2}\] For $\frac{x}{2}$ to be an integer, $x$ must be even, but not necessarily divisible by $3$. Thus, the answer is $\mathrm{(B)}$.

See also

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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
2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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

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