2024 AMC 10B Problems/Problem 6

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.

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

Solution 3

Denote the numbers as $x, \frac{2024}{x}$ . We know that per AM-GM, $\frac{x + \frac{2024}{x}}{2} <= \sqrt{x\times\frac{2024}{x}$ (Error compiling LaTeX. Unknown error_msg), so $x+\frac{2024}{x} <= 2\sqrt{2024}. Since$ \sqrt{2025} $is 45, and$ \sqrt{2024} $is slightly less than 45,$ x+\frac{2024}{x} $must be slightly less than 90, so$ 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

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

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

~ Pi Academy

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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