Difference between revisions of "2021 Fall AMC 10B Problems/Problem 5"
(Created page with "==Problem 5== 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...") |
m (→Problem 5) |
||
| Line 4: | Line 4: | ||
<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== | ||
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)^{2022}=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 | ||
Revision as of 21:54, 22 November 2021
Problem 5
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/1/d/e/1deabf4c5bf5929697341b136c546025cd6b7130.png)
Therefore, ![\[\frac{n}4=\boxed{(\textbf{E})\:4^{3032}}\]](http://latex.artofproblemsolving.com/c/b/d/cbd83381e489d51262e098ec63aed3e5d6ca9e36.png)
~kingofpineapplz
