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2024 AMC 10B Problem 6

Problem 6 of 25EasierAlgebraGeometryNumber Theory

A rectangle has integer side lengths and an area of 2024.2024. What is the least possible perimeter of the rectangle?

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Solution

The perimeter 2(+w)2(\ell + w) with w=2024\ell w = 2024 is smallest when \ell and ww are as close together as possible. Factor 2024=231123.2024 = 2^3 \cdot 11 \cdot 23. The divisor pair nearest 202445\sqrt{2024} \approx 45 is 44×46,44 \times 46, which gives perimeter 2(44+46)=180.2(44 + 46) = 180. Therefore, the answer is B.

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Concepts: factor · perimeter · optimization

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.