2024 AMC 10B Problems/Problem 6
Contents
[hide]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
Note: The square of any number ending in , written in the format where is a positive integer, is equal to .
~Cattycute
Solution 3 - AM-GM Inequality
Denote the numbers as . We know that per AM-GM, , but since , must be slightly less than 90, so 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
-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.
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:
~BenjaminDong01
Solution 5 - Get Lucky
Note: This is what I did.
Assuming it's a square,
~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
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.