Difference between revisions of "2022 AMC 12A Problems/Problem 8"
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<math>\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} ... = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}</math> | <math>\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} ... = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}</math> | ||
| − | Thus, our answer is <math>10 ^ \frac{1}{2} = \boxed{(A) \sqrt{10}}</math> | + | Thus, our answer is <math>10 ^ \frac{1}{2} = \boxed{\textbf{(A) }\sqrt{10}}</math>. |
- phuang1024 | - phuang1024 | ||
Revision as of 00:27, 12 November 2022
Problem
The infinite product
![\[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \ldots\]](http://latex.artofproblemsolving.com/3/5/1/351d2e676280300007e4b9e7dd3296d6779dd7c5.png)
evaluates to a real number. What is that number?
![$\textbf{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt[3]{100}\qquad\textbf{(C) }\sqrt[4]{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt[3]{10}$](http://latex.artofproblemsolving.com/5/8/6/58669832b19a56c4b8fda00b4c2151cba3f003f4.png)
Solution
We can write ![$\sqrt[3]{10}$](http://latex.artofproblemsolving.com/9/a/4/9a414a76d3daf242ab82d4069a47d6a9b39428da.png)
as 
. Similarly, ![$\sqrt[3]{\sqrt[3]{10}} = (10 ^ \frac{1}{3}) ^ \frac{1}{3} = 10 ^ \frac{1}{3^2}$](http://latex.artofproblemsolving.com/e/8/7/e8797d24c656db81f5590b6dd1cd96299a42f87c.png)
.
By continuing this, we get the form
which is
.
Using the formula for an infinite geometric series 
, we get
Thus, our answer is 
.
- phuang1024
See Also
| 2022 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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. 
