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

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To be administered in November 2024 and added later.
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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) } 180 \qquad\textbf{(C) } 222 \qquad\textbf{(D) } 228 \qquad\textbf{(E) } 390</math>
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 +
==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>
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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]].
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==Solution 2 - Squared Numbers Trick==
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We know that <math>x^2 = (x-1)(x+1)+1</math> . Recall that <math>45^2 = 2025</math>.
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If I want 1 less than 2025, which is 2024, I can take 1 number higher and 1 number lower from 45, which are 46 and 44. These are the 2 sides of the minimum perimeter because the 2 numbers are closest to each other, which is what we want to get the minimum.
 +
 
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Finding the perimeter with <math>2(46+44)</math> we get <math>\boxed{\textbf{(B) }180}.</math>
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Solution by ~Taha Jazaeri
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Note: The square of any positive integer with units digit <math>5</math>, written in the format <math>10x + 5</math> where <math>x</math> is a positive integer, is equal to <math>100(x)(x+1)+25</math>.
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~Cattycute
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==Solution 3 - AM-GM Inequality==
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Denote the numbers as <math>x, \frac{2024}{x}</math>. We know that per AM-GM, <math>x+\frac{2024}{x} \geq 2\sqrt{2024}</math>, but since <math>2\sqrt{2025} = 90</math>, <math>2\sqrt{2024}</math> must be slightly less than 90, so <math>2x + 2\frac{2024}{x}</math> must be slightly less than 180, eliminating A as a possible answer choice. Proceed with the following solutions above to get 44 and 46, which is <math>\boxed{\textbf{(B) }180}.</math>
 +
 
 +
-aleyang
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==Solution 4 - Difference of Squares==
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Note that the year 2025 is a perfect square. The beauty of this year shines down to 2024, which is 1 year lower. <math>45^2-1^2=2024=(45-1)(45+1)</math>
 +
 
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Knowing this is the closest possible we can get to a square with only integer factors of 2024, these two are our happy numbers! Add them up to get:
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<math>44+46+44+46=\boxed{\textbf{180}}</math>
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~BenjaminDong01
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==Solution 5 - Get Lucky==
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Note: This is what I did.
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<math>\sqrt{2025}=45</math>
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Assuming it's a square,
 +
 
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<math>45\cdot4=\boxed{\textbf{180}}</math>
 +
 
 +
~BenjaminDong01
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 +
==🎥✨ Video Solution by Scholars Foundation ➡️ Easy-to-Understand 💡✔️==
 +
 
 +
https://youtu.be/T_QESWAKUUk?si=aCvtmf24mzh2HPCJ
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 +
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
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https://youtu.be/QLziG_2e7CY?feature=shared
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 +
~ Pi Academy
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==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=24EZaeAThuE
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==Video Solution by Daily Dose of Math==
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https://youtu.be/k1bGPUrhYE4
 +
 
 +
~Thesmartgreekmathdude
 +
 
 +
==See also==
 +
{{AMC10 box|year=2024|ab=B|num-b=5|num-a=7}}
 +
{{MAA Notice}}

Latest revision as of 00:01, 22 November 2024

Problem

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

$\textbf{(A) } 160 \qquad\textbf{(B) } 180 \qquad\textbf{(C) } 222 \qquad\textbf{(D) } 228 \qquad\textbf{(E) } 390$

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.

Solution 2 - Squared Numbers Trick

We know that $x^2 = (x-1)(x+1)+1$ . Recall that $45^2 = 2025$.

If I want 1 less than 2025, which is 2024, I can take 1 number higher and 1 number lower from 45, which are 46 and 44. These are the 2 sides of the minimum perimeter because the 2 numbers are closest to each other, which is what we want to get the minimum.

Finding the perimeter with $2(46+44)$ we get $\boxed{\textbf{(B) }180}.$

Solution by ~Taha Jazaeri

Note: The square of any positive integer with units digit $5$, written in the format $10x + 5$ where $x$ is a positive integer, is equal to $100(x)(x+1)+25$.

~Cattycute

Solution 3 - AM-GM Inequality

Denote the numbers as $x, \frac{2024}{x}$. We know that per AM-GM, $x+\frac{2024}{x} \geq 2\sqrt{2024}$, but since $2\sqrt{2025} = 90$, $2\sqrt{2024}$ must be slightly less than 90, so $2x + 2\frac{2024}{x}$ must be slightly less than 180, eliminating A as a possible answer choice. Proceed with the following solutions above to get 44 and 46, which is $\boxed{\textbf{(B) }180}.$

-aleyang

Solution 4 - Difference of Squares

Note that the year 2025 is a perfect square. The beauty of this year shines down to 2024, which is 1 year lower. $45^2-1^2=2024=(45-1)(45+1)$

Knowing this is the closest possible we can get to a square with only integer factors of 2024, these two are our happy numbers! Add them up to get:

$44+46+44+46=\boxed{\textbf{180}}$ ~BenjaminDong01

Solution 5 - Get Lucky

Note: This is what I did.

$\sqrt{2025}=45$

Assuming it's a square,

$45\cdot4=\boxed{\textbf{180}}$

~BenjaminDong01

🎥✨ Video Solution by Scholars Foundation ➡️ Easy-to-Understand 💡✔️

https://youtu.be/T_QESWAKUUk?si=aCvtmf24mzh2HPCJ

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/QLziG_2e7CY?feature=shared

~ Pi Academy

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=24EZaeAThuE

Video Solution by Daily Dose of Math

https://youtu.be/k1bGPUrhYE4

~Thesmartgreekmathdude

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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