2003 AMC 12B Problem 1

Attempt Problem 1 of the 2003 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

1.

Which of the following is the same as
2−4+6−8+10−12+143−6+9−12+15−18+21?\frac{2 - 4 + 6 - 8 + 10 - 12 + 14}{3 - 6 + 9 - 12 + 15 - 18 + 21}?
−1-1
−23-\dfrac{2}{3}
23\dfrac{2}{3}
11
143\dfrac{14}{3}
Answer: C
Concepts:factoringfraction
Small hint:
Factor a common constant out of the numerator and the denominator
Big hint:
The two parenthesized sums 1−2+3−⋯+71 - 2 + 3 - \cdots + 7 are identical
Solution:
Factor 22 from the numerator and 33 from the denominator: 2(1−2+3−4+5−6+7)3(1−2+3−4+5−6+7).\frac{2(1 - 2 + 3 - 4 + 5 - 6 + 7)}{3(1 - 2 + 3 - 4 + 5 - 6 + 7)}.
The equal parenthesized sums cancel, leaving 23.\dfrac{2}{3}.
Thus, the correct answer is C.
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