2024 AMC 10B Problems/Problem 6

Problem

A rectangle has integer length sides and an area of 2024. What is the least possible perimeter of the rectangle?

$\textbf{(A) } 160 \qquad\textbf{(B) } 180 \qquad\textbf{(C) } 16 \qquad\textbf{(D) } 17 \qquad\textbf{(E) } 18$

$2024 = 44 \cdot 46$ , $\sqrt{44 + 46} \cdot 2$ = $180$ , so the solution is $\boxed{\textbf{(B) }180}$

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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