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

Revision as of 04:19, 5 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.

Let $n=8^{2022}$ . Which of the following is equal to $\frac{n}{4}?$

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

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

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

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

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-{{duplicate|[[2021 Fall AMC 10B Problems#Problem 5|2021 Fall AMC 10B #5]] and [[2021 Fall AMC 12B Problems#Problem 4|2021 Fall AMC 12B #4]]}}
+{{duplicate|[[2021 Fall AMC 10B Problems/Problem 5|2021 Fall AMC 10B #5]] and [[2021 Fall AMC 12B Problems/Problem 4|2021 Fall AMC 12B #4]]}}
 == Problem ==
 Let <math>n=8^{2022}</math>. Which of the following is equal to <math>\frac{n}{4}?</math>