Difference between revisions of "2021 Fall AMC 12B Problems/Problem 4"
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<math>n = 8^{2022} = (2^{3})^{2022} = 2^{6064} = (4^{\frac{1}{2}})^{6064} = 4^{\frac{6064}{2}} = \boxed{\textbf{(E)} \: 4^{3032}}.</math> | <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 | + | - abed_nadir (youtube.com/@indianmathguy) |
==Video Solution by Interstigation== | ==Video Solution by Interstigation== |
Latest revision as of 04:27, 12 November 2023
- 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.
Contents
Problem
Let . Which of the following is equal to
Solution 1
We have
Therefore,
~kingofpineapplz
Solution 2
The requested value is
~NH14
Solution 3
If we rewrite everything in powers of 2, we get:
- abed_nadir (youtube.com/@indianmathguy)
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=429
Video Solution (Just 3 min!)
~Education, the Study of Everything
Video Solution by WhyMath
~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.