2003 AMC 10A Problem 1 — Solution | LIVE by Po-Shen Loh

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2003 AMC 10A Problem 1

Attempt Problem 1 of the 2003 AMC 10A 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 10A solutions, or check the answer key.

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

1.

What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?

$0$

$1$

$2$

$2003$

$4006$

Answer: D

Concepts:summation pairing and grouping

Difficulty rating: 860

Solution:

The $k$ th even number $2k$ is exactly $1$ more than the $k$ th odd number $2k-1.$

Summing this difference over all $2003$ pairs gives $2003 \cdot 1 = 2003.$

Thus, the correct answer is D.

Problem 2#2 →

Problem 1 in Other Years

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