Difference between revisions of "2008 AMC 12A Problems/Problem 5"
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==Problem== | ==Problem== | ||
Suppose that | Suppose that | ||
− | + | <cmath>\frac{2x}{3}-\frac{x}{6}</cmath> | |
− | < | ||
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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>? | ||
− | <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> |
− | \ | ||
− | \ | ||
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− | ==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|num-b=4|num-a=6|ab=A}} |
Revision as of 23:46, 25 April 2008
Problem
Suppose that
is an integer. Which of the following statements must be true about
?
Solution
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 |