# 2003 AMC 12B Problems/Problem 1

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 $$\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?$$ $\text {(A) } -1 \qquad \text {(B) } -\frac{2}{3} \qquad \text {(C) } \frac{2}{3} \qquad \text {(D) } 1 \qquad \text {(E) } \frac{14}{3}$

## Solution

The numbers in the numerator and denominator can be grouped like this: \begin{align*} 2+(-4+6)+(-8+10)+(-12+14)&=2*4\\ 3+(-6+9)+(-12+15)+(-18+21)&=3*4\\ \frac{2*4}{3*4}&=\frac{2}{3} \Rightarrow \text {(C)} \end{align*}

Alternatively, notice that each term in the numerator is $\frac{2}{3}$ of a term in the denominator, so the quotient has to be $\frac{2}{3}$.

## See also

 2003 AMC 12B (Problems • Answer Key • Resources) Preceded byFirst Question Followed byProblem 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 byFirst Question Followed byProblem 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

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