2021 Fall AMC 12B Problems/Problem 4: Difference between revisions
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{{duplicate|[[2021 Fall AMC 10B Problems | {{duplicate|[[2021 Fall AMC 10B Problems/Problem 5|2021 Fall AMC 10B #5]] and [[2021 Fall AMC 12B Problems/Problem 4|2021 Fall AMC 12B #4]]}} | ||
== Problem == | == Problem == | ||
Let <math>n=8^{2022}</math>. Which of the following is equal to <math>\frac{n}{4}?</math> | Let <math>n=8^{2022}</math>. Which of the following is equal to <math>\frac{n}{4}?</math> | ||
<math> | <math>\textbf{(A)}\: 4^{1010}\qquad\textbf{(B)} \: 2^{2022}\qquad\textbf{(C)} \: 8^{2018}\qquad\textbf{(D)} \: 4^{3031}\qquad\textbf{(E)} \: 4^{3032}</math> | ||
==Solution 1== | ==Solution 1== | ||
We have <cmath>n=8^{2022}= \left(8^\frac{2}{3}\right)^{ | We have <cmath>n=8^{2022}= \left(8^\frac{2}{3}\right)^{3033}=4^{3033}.</cmath> | ||
Therefore, <cmath>\frac{n}4=\boxed{ | Therefore, <cmath>\frac{n}4=\boxed{\textbf{(E)} \: 4^{3032}}.</cmath> | ||
~kingofpineapplz | ~kingofpineapplz | ||
==Solution 2== | ==Solution 2== | ||
The requested value is <cmath>\frac{8^{2022}}{4} = \frac{2^{6066}}{4} = \frac{2^{6066}}{2^2} = 2^{6064} = | The requested value is <cmath>\frac{8^{2022}}{4} = \frac{2^{6066}}{4} = \frac{2^{6066}}{2^2} = 2^{6064} = \boxed{\textbf{(E)} \: 4^{3032}}.</cmath> | ||
~NH14 | |||
~ | ==Solution 3== | ||
If we rewrite everything in powers of 2, we get: | |||
<math>n = 8^{2022} = (2^{3})^{2022} = 2^{6064} = (4^{\frac{1}{2}})^{6064} = 4^{\frac{6064}{2}} = \boxed{\textbf{(E)} \: 4^{3032}}.</math> | |||
- abed_nadir (youtube.com/@indianmathguy) | |||
==Solution 4== | |||
<math>n = 8^{2022}</math> | |||
<math>{8^{2022}}/{4} = 2 * 8^{2021} = 2 * 2^{6063} = 2^{6064} = \boxed{\textbf{(E)} \: 4^{3032}}.</math> | |||
-sphericalfox | |||
==Video Solution by Interstigation== | |||
https://youtu.be/p9_RH4s-kBA?t=429 | |||
==Video Solution (Just 3 min!)== | |||
https://youtu.be/480KnrVnbOc | |||
~Education, the Study of Everything | |||
==Video Solution by WhyMath== | |||
https://youtu.be/iXX4WtMKU_g | |||
~savannahsolver | |||
==Video Solution by TheBeautyofMath== | |||
For AMC 10: https://youtu.be/lC7naDZ1Eu4?t=882 | |||
For AMC 12: https://youtu.be/yaE5aAmeesc?t=590 | |||
~IceMatrix | |||
==See Also== | |||
{{AMC10 box|year=2021 Fall|ab=B|num-a=6|num-b=4}} | |||
{{AMC12 box|year=2021 Fall|ab=B|num-a=5|num-b=3}} | |||
[[Category:Introductory Algebra Problems]] | |||
{{MAA Notice}} | |||
Latest revision as of 17:01, 4 November 2024
- The following problem is from both the 2021 Fall AMC 10B #5 and 2021 Fall AMC 12B #4, so both problems redirect to this page.
Problem
Let 
. Which of the following is equal to 
Solution 1
We have ![\[n=8^{2022}= \left(8^\frac{2}{3}\right)^{3033}=4^{3033}.\]](http://latex.artofproblemsolving.com/f/f/8/ff8f511842bf3275a1d48df6d5f8adcd18102756.png)
Therefore, ![\[\frac{n}4=\boxed{\textbf{(E)} \: 4^{3032}}.\]](http://latex.artofproblemsolving.com/4/4/c/44cf34fd04107d751424dd4df3f40daae9b744e1.png)
~kingofpineapplz
Solution 2
The requested value is ![\[\frac{8^{2022}}{4} = \frac{2^{6066}}{4} = \frac{2^{6066}}{2^2} = 2^{6064} = \boxed{\textbf{(E)} \: 4^{3032}}.\]](http://latex.artofproblemsolving.com/7/d/8/7d81ce2dc12bcdcd5021c0d085e79c9f61a4ce52.png)
~NH14
Solution 3
If we rewrite everything in powers of 2, we get:
- abed_nadir (youtube.com/@indianmathguy)
Solution 4
-sphericalfox
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=429
Video Solution (Just 3 min!)
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
For AMC 10: https://youtu.be/lC7naDZ1Eu4?t=882
For AMC 12: https://youtu.be/yaE5aAmeesc?t=590
~IceMatrix
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
| 2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America. 
