Difference between revisions of "2024 AMC 10B Problems/Problem 6"
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==Solution 1== | ==Solution 1== | ||
| − | + | <math>2024 = 44 \cdot 46</math>, <math>sqrt(44 + 46) * 2</math> = <math>180</math>, so the solution is <math>\boxed{\textbf{(B) }180}</math> | |
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=B|num-b=5|num-a=7}} | {{AMC10 box|year=2024|ab=B|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 01:02, 14 November 2024
Problem
A rectangle has integer length sides and an area of 2024. What is the least possible perimeter of the rectangle?
Solution 1
, 
= 
, so the solution is 
See also
| 2024 AMC 10B (Problems • Answer Key • Resources) | ||
| 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. 
