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

Revision as of 11:35, 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.

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

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
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.

@@ Line 30: / Line 30: @@
 Finding the perimeter with <math>2(46+44)</math> we get <math>\boxed{\textbf{(B) }180}.</math>
-Solution by ~Taha Jazaeri & Sri Sambhara
+Solution by ~Taha Jazaeri
 ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==

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