Difference between revisions of "2024 AMC 10B Problems/Problem 6"
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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) } | + | <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 - Prime Factorization== |
| − | + | We can start by assigning the values x and y for both sides. Here is the equation representing the area: | |
| + | |||
| + | |||
| + | <math>x \cdot y = 2024</math> | ||
| + | |||
| + | Let's write out 2024 fully factorized. | ||
| + | |||
| + | |||
| + | <math>2^3 \cdot 11 \cdot 23</math> | ||
| + | |||
| + | Since we know that <math>x^2 > (x+1)(x-1)</math>, we want the two closest numbers possible. After some quick analysis, those two numbers are <math>44</math> and <math>46</math>. <math>\\44+46=90</math> | ||
| + | |||
| + | Now we multiply by <math>2</math> and get <math>\boxed{\textbf{(B) }180}.</math> | ||
| + | |||
| + | Solution by [[User:IshikaSaini|IshikaSaini]]. | ||
==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}} | ||
Latest revision as of 01:30, 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 - Prime Factorization
We can start by assigning the values x and y for both sides. Here is the equation representing the area:
Let's write out 2024 fully factorized.
Since we know that 
, we want the two closest numbers possible. After some quick analysis, those two numbers are 
and 
. 
Now we multiply by 
and get 
Solution by IshikaSaini.
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. 
