# 2021 Fall AMC 12B Problems/Problem 4

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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 (Just 3 min!)

~Education, the Study of Everything

~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 byProblem 4 Followed byProblem 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 byProblem 3 Followed byProblem 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.