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

m (Solution 1)
m (Problem)
Line 2: Line 2:
 
A rectangle has integer length sides and an area of 2024. What is the least possible perimeter of the rectangle?
 
A rectangle has integer length sides and an area of 2024. What is the least possible perimeter of the rectangle?
  
<math>\textbf{(A) } 160 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 16 \qquad\textbf{(D) } 17 \qquad\textbf{(E) } 18</math>
+
<math>\textbf{(A) } 160 \qquad\textbf{(B) } 180 \qquad\textbf{(C) } 16 \qquad\textbf{(D) } 17 \qquad\textbf{(E) } 18</math>
  
 
==Solution 1==
 
==Solution 1==

Revision as of 01:06, 14 November 2024

Problem

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

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

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png