2003 AMC 10B Problem 1

The numerator is $2 \cdot 1 - 2 \cdot 2 + 2 \cdot 3 - \cdots + 2 \cdot 7$ and the denominator is $3 \cdot 1 - 3 \cdot 2 + 3 \cdot 3 - \cdots + 3 \cdot 7$. Factor out the multipliers:
$$\frac{2\,(1-2+3-4+5-6+7)}{3\,(1-2+3-4+5-6+7)}.$$

The alternating sum $1-2+3-4+5-6+7$ equals $(1-2)+(3-4)+(5-6)+7 = 4$, which is not zero, so it cancels and the fraction equals $\frac{2}{3}$.

As a check, the numerator is $(2-4)+(6-8)+(10-12)+14 = 8$ and the denominator is $(3-6)+(9-12)+(15-18)+21 = 12$, and $\frac{8}{12} = \frac{2}{3}$.

The answer is $\boxed{\textbf{(C)}\ \frac{2}{3}}$.

Both rows are the same alternating sum scaled by different constants, and scaling a sum scales its value. Whenever two expressions are built from proportional pieces, look for the common factor before doing any arithmetic; it turns a seven-term computation into a one-line cancellation.

• Dropping a sign while adding and getting $-\frac{2}{3}$, or dividing the last numerator term by the first denominator term to get $\frac{14}{3}$.
• Pairing the terms as $(2-4)+(6-8)+(10-12)$ and forgetting the unpaired final term $14$.