Difference between revisions of "2021 Fall AMC 12B Problems/Problem 4"
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| − | {{AMC10 box|year=2021 Fall|ab=B|num-a= | + | {{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}} | {{AMC12 box|year=2021 Fall|ab=B|num-a=5|num-b=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 01:21, 24 November 2021
- 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 ![\[n=8^{2022}= \left(8^\frac{2}{3}\right)^{2022}=4^{3033}.\]](http://latex.artofproblemsolving.com/6/b/f/6bff93b33bc90f12c694a4edf3d2587b9aab71c8.png)
Therefore, ![\[\frac{n}4=\boxed{(\textbf{E})\:4^{3032}}.\]](http://latex.artofproblemsolving.com/3/3/b/33badf2380db5fd9beb44c3dcc25766267038e1e.png)
~kingofpineapplz
Solution 2
The requested value is ![\[\frac{8^{2022}}{4} = \frac{2^{6066}}{4} = \frac{2^{6066}}{2^2} = 2^{6064} = 4^{3032}.\]](http://latex.artofproblemsolving.com/a/b/d/abdf15b3996c95b132a519f8839ccb05b8f979a9.png)
Thus, the answer is 
~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. 
