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2024 AMC 10B Problem 6

Problem 6 of 25EasierGeometryNumber TheoryProblem-Solving Techniques

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(ℓ+w)2(\ell + w) with ℓw=2024\ell w = 2024 is smallest when ℓ\ell and ww are as close together as possible. Factor 2024=23⋅11⋅23.2024 = 2^3 \cdot 11 \cdot 23. The divisor pair nearest 2024≈45\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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Tagged: factor · perimeter · optimization

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