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Difference between revisions of "2008 AMC 12A Problems/Problem 5"

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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