Difference between revisions of "2021 Fall AMC 12B Problems/Problem 4"

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~savannahsolver
 
~savannahsolver
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==Video Solution by TheBeautyofMath==
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For AMC 10: https://youtu.be/lC7naDZ1Eu4?t=882
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For AMC 12: https://youtu.be/yaE5aAmeesc?t=590
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~IceMatrix
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=6|num-b=4}}
 
{{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 00:31, 30 December 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 $n=8^{2022}$. Which of the following is equal to $\frac{n}{4}?$

$\textbf{(A)}\: 4^{1010}\qquad\textbf{(B)} \: 2^{2022}\qquad\textbf{(C)} \: 8^{2018}\qquad\textbf{(D)} \: 4^{3031}\qquad\textbf{(E)} \: 4^{3032}$

Solution 1

We have \[n=8^{2022}=  \left(8^\frac{2}{3}\right)^{3033}=4^{3033}.\] Therefore, \[\frac{n}4=\boxed{\textbf{(E)} \: 4^{3032}}.\] ~kingofpineapplz

Solution 2

The requested value is \[\frac{8^{2022}}{4} = \frac{2^{6066}}{4} = \frac{2^{6066}}{2^2} = 2^{6064} = \boxed{\textbf{(E)} \: 4^{3032}}.\] ~NH14

Video Solution by Interstigation

https://youtu.be/p9_RH4s-kBA?t=429

Video Solution

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

2021 Fall AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png