Difference between revisions of "2021 Fall AMC 12B Problems/Problem 4"
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==Solution 1== | ==Solution 1== | ||
We have <cmath>n=8^{2022}= \left(8^\frac{2}{3}\right)^{3033}=4^{3033}.</cmath> | 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> |
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~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> |
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~NH14 | ~NH14 | ||
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==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
Revision as of 12:16, 4 January 2022
- 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
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=429
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
