Difference between revisions of "2008 AMC 12A Problems/Problem 5"
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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> | |
| − | < | ||
| − | |||
| − | |||
| − | |||
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>\ | + | <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== |
| − | + | <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 | + | ==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
![\[\frac{2x}{3}-\frac{x}{6}\]](http://latex.artofproblemsolving.com/5/0/8/5086ca4e08ddc4e7a20b6b10fdf8d0824706658f.png)
is an integer. Which of the following statements must be true about 
?
Solution
![\[\frac{2x}{3}-\frac{x}{6}\quad\Longrightarrow\quad\frac{4x}{6}-\frac{x}{6}\quad\Longrightarrow\quad\frac{3x}{6}\quad\Longrightarrow\quad\frac{x}{2}\]](http://latex.artofproblemsolving.com/6/b/f/6bfcc085910df27c7cc1f6881760996e40ed36ff.png)
For 
to be an integer, must be even, but not necessarily divisible by 
. Thus, the answer is 
.
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. 
