2024 AMC 10B Problems/Problem 6: Difference between revisions
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Finding the perimeter with <math>2(46+44)</math> we get <math>\boxed{\textbf{(B) }180}.</math> | Finding the perimeter with <math>2(46+44)</math> we get <math>\boxed{\textbf{(B) }180}.</math> | ||
Solution by ~Taha Jazaeri | Solution by ~Taha Jazaeri | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ||
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?
Solution 1 - Prime Factorization
We can start by assigning the values x and y for both sides. Here is the equation representing the area:
Let's write out 2024 fully factorized.
Since we know that 
, we want the two closest numbers possible. After some quick analysis, those two numbers are 
and 
. 
Now we multiply by 
and get 
Solution by IshikaSaini.
Solution 2 - Squared Numbers Trick
We know that 
. Recall that 
.
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 
we get
Solution by ~Taha Jazaeri
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/QLziG_2e7CY?feature=shared
~ Pi Academy
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 | ||
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