Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
 
(One intermediate revision by one other user not shown)
Line 5: Line 5:
 
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>\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>
+
<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==
Line 14: Line 14:
 
{{AMC12 box|year=2008|ab=A|num-b=4|num-a=6}}
 
{{AMC12 box|year=2008|ab=A|num-b=4|num-a=6}}
 
{{AMC10 box|year=2008|ab=A|num-b=8|num-a=10}}
 
{{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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_5&oldid=159625"