Difference between revisions of "2022 AMC 12A Problems/Problem 8"
MRENTHUSIASM (talk | contribs) m (→Solution 1) |
|||
| Line 40: | Line 40: | ||
~kxiang | ~kxiang | ||
| − | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | + | ==Video Solution 1 (HOW TO THINK CREATIVELY!!!)== |
https://youtu.be/_YDTIEuXTzY | https://youtu.be/_YDTIEuXTzY | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution 2 (Smart and Fun!!!)== | ||
| + | https://youtu.be/7yAh4MtJ8a8?si=OJHbJh4_xMjBc9OY&t=1397 | ||
| + | |||
| + | ~Math-X | ||
| + | |||
== See Also == | == See Also == | ||
Revision as of 11:34, 25 October 2023
Contents
Problem
The infinite product
![\[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\]](http://latex.artofproblemsolving.com/1/a/1/1a14175298870f69d36ba1659437da6d34b31ae3.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 1
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
![\[10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \cdots,\]](http://latex.artofproblemsolving.com/d/7/d/d7d5fb0d94d20028cb62feaa5265dab268fa4e78.png)
which is
![\[10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots}.\]](http://latex.artofproblemsolving.com/4/b/a/4ba313b8a301854a619c61e4095435f11f64645a.png)
Using the formula for an infinite geometric series 
, we get
![\[\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}.\]](http://latex.artofproblemsolving.com/e/3/f/e3fb02e6bcb588355f1debc78a632d25064322cd.png)
Thus, our answer is 
.
- phuang1024
Solution 2
We can write this infinite product as 
(we know from the answer choices that the product must converge):
![\[L = \sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots.\]](http://latex.artofproblemsolving.com/c/5/d/c5ded3e8663e3424f44372047e8da32737150ac3.png)
If we raise everything to the third power, we get:
![\[L^3 = 10 \, \cdot \, \sqrt[3]{10} \, \cdot \, \sqrt[3]{\sqrt[3]{10}} \cdots = 10L \implies L^3 - 10L = 0 \implies L \in \left\{0, \pm \sqrt{10}\right\}.\]](http://latex.artofproblemsolving.com/a/8/1/a8144495c9ae8d630eb8ced96e3cc89a0800f4c2.png)
Since is positive (as it is an infinite product of positive numbers), it must be that 
~ Oxymoronic15
Solution 3
Move the first term inside the second radical. We get
![\[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots = \sqrt[3]{10\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots.\]](http://latex.artofproblemsolving.com/d/7/8/d78877d15433a45a0711b1d7dd3bb16ed5fa6997.png)
Do this for the third radical as well:
![\[\sqrt[3]{10\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots = \sqrt[3]{10\sqrt[3]{10}\sqrt[3]{\sqrt[3]{10}}} \cdots = \sqrt[3]{10\sqrt[3]{10\sqrt[3]{10\cdots}}}.\]](http://latex.artofproblemsolving.com/5/c/d/5cd6d600e2e529810b0c095e3472f714946e4ddd.png)
It is clear what the pattern is. Setting the answer as 
we have ![\[P = \sqrt[3]{10P},\]](http://latex.artofproblemsolving.com/1/1/2/112b34b1d18f0849184e48a496cfd0b7d8bfb271.png)
from which 
~kxiang
Video Solution 1 (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
Video Solution 2 (Smart and Fun!!!)
https://youtu.be/7yAh4MtJ8a8?si=OJHbJh4_xMjBc9OY&t=1397
~Math-X
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. 
