2024 AMC 10B Problems/Problem 6

Revision as of 01:02, 14 November 2024 by Ishikasaini (talk | contribs) (Solution 1)

Problem

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

$\textbf{(A) } 160 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 16 \qquad\textbf{(D) } 17 \qquad\textbf{(E) } 18$

Solution 1

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

See also

2024 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AMC 10 Problems and Solutions

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