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2003 AMC 12B Problem 1

Below is the professionally curated solution for Problem 1 of the 2003 AMC 12B, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:factoringfraction

Difficulty rating: 840

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}

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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