Difference between revisions of "2003 AMC 12B Problems/Problem 1"
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==Solution== | ==Solution== | ||
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| + | The numbers in the numerator and denominator can be grouped like this: | ||
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<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
| − | 2-4+6-8+10-12+14= | + | 2+(-4+6)+(-8+10)+(-12+14)&=2*4\\ |
| − | 3-6+9-12+15-18+21= | + | 3+(-6+9)+(-12+15)+(-18+21)&=3*4\\ |
| − | \frac{2 | + | \frac{2*4}{3*4}&=\frac{2}{3} \Rightarrow \text {(C)} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Latest revision as of 23:09, 13 September 2015
- The following problem is from both the 2003 AMC 12B #1 and 2003 AMC 10B #1, so both problems redirect to this page.
Problem
Which of the following is the same as
![\[\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?\]](http://latex.artofproblemsolving.com/d/d/a/ddaa33db347793c4318c6123185c74195fe22c74.png)
Solution
The numbers in the numerator and denominator can be grouped like this:
Alternatively, notice that each term in the numerator is 
of a term in the denominator, so the quotient has to be .
See also
| 2003 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by First Question |
Followed by Problem 2 |
| 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 | |
| 2003 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 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. 
