Difference between revisions of "2024 AMC 10B Problems/Problem 6"

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==Problem==
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A rectangle has integer length sides and an area of 2024. What is the least possible perimeter of the rectangle?
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<math>\textbf{(A) } 160 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 16 \qquad\textbf{(D) } 17 \qquad\textbf{(E) } 18</math>
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==Solution 1==
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B) 180, 2024=44*46, (44+46)*2=180.
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==See also==
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{{AMC10 box|year=2024|ab=B|num-b=5|num-a=7}}
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{{MAA Notice}}

Revision as of 00:41, 14 November 2024

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

B) 180, 2024=44*46, (44+46)*2=180.

See also

2024 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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