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

~kingofpineapplz

Solution 2

The requested value is $\frac{8^{2022}}{4} = \frac{2^{6066}}{4} = \frac{2^{6066}}{2^2} = 2^{6064} = 4^{3032}.$ Thus, the answer is $\boxed{(\textbf{E}) \: 4^{3032}}.$

~NH14

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2021 Fall AMC 12B Problems/Problem 4

Problem

Solution 1

Solution 2