Difference between revisions of "2024 AMC 10B Problems/Problem 6"
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Solution by ~Taha Jazaeri | Solution by ~Taha Jazaeri | ||
| − | ==Solution 3== | + | Note: The square of any positive integer with units digit <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>. |
| − | Denote the numbers as <math>x, \frac{2024}{x}</math>. We know that per AM-GM, <math>x+\frac{2024}{x} | + | |
| + | ~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} \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 | -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. <math>45^2-1^2=2024=(45-1)(45+1)</math> | ||
| + | |||
| + | 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: | ||
| + | |||
| + | <math>44+46+44+46=\boxed{\textbf{180}}</math> | ||
| + | ~BenjaminDong01 | ||
| + | |||
| + | ==Solution 5 - Get Lucky== | ||
| + | Note: This is what I did. | ||
| + | |||
| + | <math>\sqrt{2025}=45</math> | ||
| + | |||
| + | Assuming it's a square, | ||
| + | |||
| + | <math>45\cdot4=\boxed{\textbf{180}}</math> | ||
| + | |||
| + | ~BenjaminDong01 | ||
| + | |||
| + | ==🎥✨ Video Solution by Scholars Foundation ➡️ Easy-to-Understand 💡✔️== | ||
| + | |||
| + | https://youtu.be/T_QESWAKUUk?si=aCvtmf24mzh2HPCJ | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ||
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==Video Solution 2 by SpreadTheMathLove== | ==Video Solution 2 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=24EZaeAThuE | https://www.youtube.com/watch?v=24EZaeAThuE | ||
| + | |||
| + | ==Video Solution by Daily Dose of Math== | ||
| + | |||
| + | https://youtu.be/k1bGPUrhYE4 | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=B|num-b=5|num-a=7}} | {{AMC10 box|year=2024|ab=B|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:01, 22 November 2024
Contents
- 1 Problem
- 2 Solution 1 - Prime Factorization
- 3 Solution 2 - Squared Numbers Trick
- 4 Solution 3 - AM-GM Inequality
- 5 Solution 4 - Difference of Squares
- 6 Solution 5 - Get Lucky
- 7 🎥✨ Video Solution by Scholars Foundation ➡️ Easy-to-Understand 💡✔️
- 8 Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
- 9 Video Solution 2 by SpreadTheMathLove
- 10 Video Solution by Daily Dose of Math
- 11 See also
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 positive integer with units digit 
, 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 by Scholars Foundation ➡️ Easy-to-Understand 💡✔️
https://youtu.be/T_QESWAKUUk?si=aCvtmf24mzh2HPCJ
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
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
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. 
