2021 Fall AMC 12B Problems/Problem 4

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

Solution 3

If we rewrite everything in powers of 2, we get: $n = 8^{2022} = (2^{3})^{2022} = 2^{6064} = (4^{\frac{1}{2}})^{6064} = 4^{\frac{6064}{2}} = \boxed{\textbf{(E)} \: 4^{3032}}.$

- abed_nadir (youtube.com/@indianmathguy)

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

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

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