1999 AMC 12 Problem 1 — Solution | LIVE by Po-Shen Loh

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1999 AMC 12 Problem 1

Below is the professionally curated solution for Problem 1 of the 1999 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 12 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:summation pairing and grouping

Difficulty rating: 800

1.

$1 - 2 + 3 - 4 + \cdots - 98 + 99 = \, ?$

$-50$

$-49$

$0$

$49$

$50$

Solution:

Pairing consecutive terms gives $(1-2) + (3-4) + \cdots + (97-98) + 99.$ There are $49$ pairs, each equal to $-1,$ so the sum is $-49 + 99 = 50.$

Thus, the correct answer is E.

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Problem 1 in Other Years

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