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

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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]]}}
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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]]}}
 
== Problem ==
 
== Problem ==
 
Let <math>n=8^{2022}</math>. Which of the following is equal to <math>\frac{n}{4}?</math>
 
Let <math>n=8^{2022}</math>. Which of the following is equal to <math>\frac{n}{4}?</math>
  
−
<math>(\textbf{A})\: 4^{1010}\qquad(\textbf{B}) \: 2^{2022}\qquad(\textbf{C}) \: 8^{2018}\qquad(\textbf{D}) \: 4^{3031}\qquad(\textbf{E}) \: 4^{3032}</math>
+
<math>\textbf{(A)}\: 4^{1010}\qquad\textbf{(B)} \: 2^{2022}\qquad\textbf{(C)} \: 8^{2018}\qquad\textbf{(D)} \: 4^{3031}\qquad\textbf{(E)} \: 4^{3032}</math>
  
 
==Solution 1==
 
==Solution 1==
−
We have <cmath>n=8^{2022}=  \left(8^\frac{2}{3}\right)^{2022}=4^{3033}.</cmath>
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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>
+
Therefore, <cmath>\frac{n}4=\boxed{\textbf{(E)} \: 4^{3032}}.</cmath>
−
 
 
 
~kingofpineapplz
 
~kingofpineapplz
  
 
==Solution 2==
 
==Solution 2==
  
−
The requested value is <cmath>\frac{8^{2022}}{4} = \frac{2^{6066}}{4} = \frac{2^{6066}}{2^2} = 2^{6064} = 4^{3032}.</cmath> Thus, the answer is <math>\boxed{(\textbf{E}) \: 4^{3032}}.</math>
+
The requested value is <cmath>\frac{8^{2022}}{4} = \frac{2^{6066}}{4} = \frac{2^{6066}}{2^2} = 2^{6064} = \boxed{\textbf{(E)} \: 4^{3032}}.</cmath>
 +
~NH14
 +
 
 +
==Solution 3==
 +
 
 +
If we rewrite everything in powers of 2, we get:
 +
<math>n = 8^{2022} = (2^{3})^{2022} = 2^{6064} = (4^{\frac{1}{2}})^{6064} = 4^{\frac{6064}{2}} = \boxed{\textbf{(E)} \: 4^{3032}}.</math>
 +
 
 +
- abed_nadir (youtube.com/@indianmathguy)
 +
 
 +
==Solution 4==
 +
<math>n = 8^{2022}</math>
 +
 
 +
<math>{8^{2022}}/{4} = 2 * 8^{2021} = 2 * 2^{6063} = 2^{6064} = \boxed{\textbf{(E)} \: 4^{3032}}.</math>
 +
 
 +
-sphericalfox
  
−
~NH14
 
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
 
https://youtu.be/p9_RH4s-kBA?t=429
 
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==
 
==See Also==
−
{{AMC10 box|year=2021 Fall|ab=B|num-a=5|num-b=3}}
+
{{AMC10 box|year=2021 Fall|ab=B|num-a=6|num-b=4}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=5|num-b=3}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=5|num-b=3}}
 +
 +
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:01, 4 November 2024

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)

Solution 4

$n = 8^{2022}$

${8^{2022}}/{4} = 2 * 8^{2021} = 2 * 2^{6063} = 2^{6064} = \boxed{\textbf{(E)} \: 4^{3032}}.$

-sphericalfox

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 (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. AMC logo.png

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