2022 AMC 12A Problems/Problem 8: Difference between revisions
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The infinite product | The infinite product | ||
<cmath>\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \ | <cmath>\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots</cmath> | ||
evaluates to a real number. What is that number? | evaluates to a real number. What is that number? | ||
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By continuing this, we get the form | By continuing this, we get the form | ||
<math>10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \ | <math>10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \cdots</math> | ||
which is | which is | ||
<math>10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + | <math>10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots}</math>. | ||
Using the formula for an infinite geometric series <math>S = \frac{a}{1-r}</math>, we get | Using the formula for an infinite geometric series <math>S = \frac{a}{1-r}</math>, we get | ||
<math>\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + | <math>\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}</math> | ||
Thus, our answer is <math>10 ^ \frac{1}{2} = \boxed{\textbf{(A) }\sqrt{10}}</math>. | Thus, our answer is <math>10 ^ \frac{1}{2} = \boxed{\textbf{(A) }\sqrt{10}}</math>. | ||
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We can write this infinite product as <math>L</math> (we know from the answer choices that the product must converge): | We can write this infinite product as <math>L</math> (we know from the answer choices that the product must converge): | ||
<cmath>L = \sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \ | <cmath>L = \sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots</cmath> | ||
If we raise everything to the <math>3^{rd}</math> power, we get: | If we raise everything to the <math>3^{rd}</math> power, we get: | ||
<cmath>L^3 = 10 \, \cdot \, \sqrt[3]{10} \, \cdot \, \sqrt[3]{\sqrt[3]{10}} \ | <cmath>L^3 = 10 \, \cdot \, \sqrt[3]{10} \, \cdot \, \sqrt[3]{\sqrt[3]{10}} \cdots = 10L \implies L^3 - 10L = 0 \implies L \in \{0, \pm \sqrt{10}\}</cmath> | ||
Since <math>L</math> is positive (it is an infinite product of positive numbers), it must be that <math>L = \boxed{\textbf{(A) }\sqrt{10}}</math>. | Since <math>L</math> is positive (it is an infinite product of positive numbers), it must be that <math>L = \boxed{\textbf{(A) }\sqrt{10}}</math>. | ||
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==Solution 3== | ==Solution 3== | ||
Move the first term inside the second radical. We get | Move the first term inside the second radical. We get | ||
<cmath>\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \ | <cmath>\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</cmath> | ||
Do this for the third radical as well. | Do this for the third radical as well. | ||
<cmath>\sqrt[3]{10\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \ | <cmath>\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}}}</cmath> | ||
It is clear what the pattern is. Setting the answer as <math>P,</math> we have <cmath>P = \sqrt[3]{10P} = \boxed{\sqrt{10}}.</cmath> | |||
~kxiang | ~kxiang | ||
Revision as of 21:27, 15 October 2023
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
which is
.
Using the formula for an infinite geometric series 
, we get
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/8/9/5/8958e230f1d0a7ca9c2dad8084a5eaf86857e355.png)
If we raise everything to the 
power, we get:
![\[L^3 = 10 \, \cdot \, \sqrt[3]{10} \, \cdot \, \sqrt[3]{\sqrt[3]{10}} \cdots = 10L \implies L^3 - 10L = 0 \implies L \in \{0, \pm \sqrt{10}\}\]](http://latex.artofproblemsolving.com/4/2/8/428fc04385cbeca3a9ab7febde7ac8d901d255fc.png)
Since is positive (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/3/4/7/3478e4b04decf204ebdfd77c0956ca0d5256314b.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/6/b/c/6bccc9667df67bda8e4441a6baa8a964ada1e66e.png)
It is clear what the pattern is. Setting the answer as 
we have ![\[P = \sqrt[3]{10P} = \boxed{\sqrt{10}}.\]](http://latex.artofproblemsolving.com/3/8/6/386095e5bbbae090e7b6cad7260386506e51029a.png)
~kxiang
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
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 | |
These problems are copyrighted © by the Mathematical Association of America. 
