2024 AMC 10B Problem 6

Below is the professionally curated solution for Problem 6 of the 2024 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 10B solutions, or check the answer key.

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Concepts:factor perimeter optimization

Difficulty rating: 1200

A rectangle has integer side lengths and an area of $2024.$ What is the least possible perimeter of the rectangle?

$160$

$180$

$222$

$228$

$390$

Solution:

The perimeter $2(\ell + w)$ with $\ell w = 2024$ is smallest when $\ell$ and $w$ are as close together as possible. Factor $2024 = 2^3 \cdot 11 \cdot 23.$ The divisor pair nearest $\sqrt{2024} \approx 45$ is $44 \times 46,$ which gives perimeter $2(44 + 46) = 180.$ Therefore, the answer is B.

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