2003 AMC 10B Problem 3

Below is the professionally curated solution for Problem 3 of the 2003 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10B solutions, or check the answer key.

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Concepts:summation linear equation

Difficulty rating: 880

The sum of $5$ consecutive even integers is $4$ less than the sum of the first $8$ consecutive odd counting numbers. What is the smallest of the even integers?

$6$

$8$

$10$

$12$

$14$

Solution:

The first $8$ odd counting numbers sum to $1+3+\cdots+15=64.$

Letting $n$ be the smallest even integer, $n+(n+2)+(n+4)+(n+6)+(n+8)=5n+20=60,$ so $n=8.$

Thus, the correct answer is B.

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