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

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<math>\textbf{(A) } 160 \qquad\textbf{(B) } 180 \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>
  
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==Solution 1==
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==Solution 1 - Prime Factorization==
  
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<math>2024 = 44 \cdot 46</math>, <math>\sqrt{44 + 46} \cdot 2</math> = <math>180</math>, so the solution is <math>\boxed{\textbf{(B) }180}</math>
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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>
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Let's write out 2024 fully factorized.
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<math>2^3 \cdot 11 \cdot 23</math>
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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>
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Now we multiply by <math>2</math> and get <math>\boxed{\textbf{(B) }180}.</math>
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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}}

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?

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

Solution 1 - Prime Factorization

We can start by assigning the values x and y for both sides. Here is the equation representing the area:


$x \cdot y = 2024$

Let's write out 2024 fully factorized.


$2^3 \cdot 11 \cdot 23$

Since we know that $x^2 > (x+1)(x-1)$, we want the two closest numbers possible. After some quick analysis, those two numbers are $44$ and $46$. $\\44+46=90$

Now we multiply by $2$ and get $\boxed{\textbf{(B) }180}.$

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. AMC logo.png

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