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

Revision as of 20:22, 31 December 2021

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

~NH14

Solution 3

We have $\begin{align*} \frac{n}{4} & = \frac{8^{2022}}{4} \\ & = \frac{2^{6066}}{4} \\ & = \frac{4^{3033}}{4} \\ & = 4^{3032} . \end{align*}$

Therefore, the answer is $\boxed{\textbf{(E) }4^{3032}}$ .

~Steven Chen (www.professorchenedu.com)

Video Solution by Interstigation

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 6: / Line 6: @@
 ==Solution 1==
-We have <cmath>n=8^{2022}=  \left(8^\frac{2}{3}\right)^{2022}=4^{3033}.</cmath>
+We have <cmath>n=8^{2022}=  \left(8^\frac{2}{3}\right)^{3033}=4^{3033}.</cmath>
 Therefore, <cmath>\frac{n}4=\boxed{(\textbf{E})\:4^{3032}}.</cmath>

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