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

Latest revision as of 18:30, 15 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 number ending in $5$ , 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 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

@@ Line 32: / Line 32: @@
 Solution by ~Taha Jazaeri
-Note: The square of any number ending in 5, 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>.
+Note: The square of any number ending in <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>.
 ~Cattycute
 ==Solution 3 - AM-GM Inequality==
-Denote the numbers as <math>x, \frac{2024}{x}</math>. We know that per AM-GM, <math>x+\frac{2024}{x} <= 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>
+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

2024 AMC 10B (Problems • Answer Key • Resources)
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