Difference between revisions of "2021 Fall AMC 12B Problems/Problem 4"
(Created page with "== Problem == 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(\tex...") |
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| + | {{duplicate|[[2021 Fall AMC 10B Problems#Problem 5|2021 Fall AMC 10B #5]] and [[2021 Fall AMC 10B 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> | ||
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<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== | + | ==Solution 1== |
| + | We have <cmath>n=8^{2022}= \left(8^\frac{2}{3}\right)^{2022}=4^{3033}.</cmath> | ||
| + | Therefore, <cmath>\frac{n}4=\boxed{(\textbf{E})\:4^{3032}}.</cmath> | ||
| + | |||
| + | ~kingofpineapplz | ||
| + | |||
| + | ==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} = 4^{3032}.</cmath> Thus, the answer is <math>\boxed{(\textbf{E}) \: 4^{3032}}.</math> | ||
~NH14 | ~NH14 | ||
Revision as of 01:17, 24 November 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 
. Which of the following is equal to 
Solution 1
We have ![\[n=8^{2022}= \left(8^\frac{2}{3}\right)^{2022}=4^{3033}.\]](http://latex.artofproblemsolving.com/6/b/f/6bff93b33bc90f12c694a4edf3d2587b9aab71c8.png)
Therefore, ![\[\frac{n}4=\boxed{(\textbf{E})\:4^{3032}}.\]](http://latex.artofproblemsolving.com/3/3/b/33badf2380db5fd9beb44c3dcc25766267038e1e.png)
~kingofpineapplz
Solution 2
The requested value is ![\[\frac{8^{2022}}{4} = \frac{2^{6066}}{4} = \frac{2^{6066}}{2^2} = 2^{6064} = 4^{3032}.\]](http://latex.artofproblemsolving.com/a/b/d/abdf15b3996c95b132a519f8839ccb05b8f979a9.png)
Thus, the answer is 
~NH14
