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Difference between revisions of "2022 AMC 12A Problems/Problem 8"
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The infinite product
 
The infinite product
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<cmath>\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \ldots</cmath>
+
<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
−
 
+
<cmath>10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \cdots,</cmath>
−
<math>10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} ...</math>
 
−
 
 
 
which is
 
which is
−
 
+
<cmath>10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots}.</cmath>
−
<math>10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} ...}</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
−
 
+
<cmath>\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}.</cmath>
−
<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{\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}}} \cdots.</cmath>
 +
If we raise everything to the third power, we get:
 +
<cmath>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\}.</cmath>
 +
Since <math>L</math> is positive (as it is an infinite product of positive numbers), it must be that <math>L = \boxed{\textbf{(A) }\sqrt{10}}.</math>
  
−
<cmath>L = \sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \ldots</cmath>
+
~ Oxymoronic15
  
−
If we raise everything to the <math>3^{rd}</math> power, we get:
+
==Solution 3==
 +
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}}} \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:
 +
<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},</cmath> from which <math>P = \boxed{\sqrt{10}}.</math>
  
−
<cmath>L^3 =  10 \, \cdot \, \sqrt[3]{10} \, \cdot \, \sqrt[3]{\sqrt[3]{10}} \ldots = 10L \implies L^3 - 10L = 0 \implies L \in \{0, \pm \sqrt{10}\}</cmath>
+
~kxiang
  
−
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>.
+
==Solution 4==
 +
Set the product equal to P. We get
 +
<cmath>\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots=P</cmath> Since this is an infinite product, there may exist a clever manipulation where we set two different espressions involving <math>P</math> equal, from which we could solve for a number. Since all terms are raised to the <math>\frac{1}{3}</math> power, we can cube both sides of our equation. We would get <cmath>10 \cdot \sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots=P^3</cmath> From this, we know that <math>\frac{P^3}{10}=P</math>, and to this equation, the only solution is <math>P = \boxed{\sqrt{10}}</math>
  
 +
~lmpofu
  
 +
==Video Solution 1 (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/_YDTIEuXTzY
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution 2 (Smart and Fun!!!)==
 +
https://youtu.be/7yAh4MtJ8a8?si=OJHbJh4_xMjBc9OY&t=1397
 +
 +
~Math-X
  
−
~ Oxymoronic15
 
  
 
== See Also ==
 
== See Also ==

Latest revision as of 14:36, 5 October 2024

Problem

The infinite product \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\] evaluates to a real number. What is that number?

$\textbf{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt[3]{100}\qquad\textbf{(C) }\sqrt[4]{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt[3]{10}$

Solution 1

We can write $\sqrt[3]{10}$ as $10 ^ \frac{1}{3}$. Similarly, $\sqrt[3]{\sqrt[3]{10}} = (10 ^ \frac{1}{3}) ^ \frac{1}{3} = 10 ^ \frac{1}{3^2}$.

By continuing this, we get the form \[10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \cdots,\] which is \[10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots}.\] Using the formula for an infinite geometric series $S = \frac{a}{1-r}$, we get \[\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}.\] Thus, our answer is $10 ^ \frac{1}{2} = \boxed{\textbf{(A) }\sqrt{10}}$.

- phuang1024

Solution 2

We can write this infinite product as $L$ (we know from the answer choices that the product must converge): \[L = \sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots.\] If we raise everything to the third power, we get: \[L^3 = 10 \, \cdot \, \sqrt[3]{10} \, \cdot \, \sqrt[3]{\sqrt[3]{10}} \cdots = 10L \implies L^3 - 10L = 0 \implies L \in \left\{0, \pm \sqrt{10}\right\}.\] Since $L$ is positive (as it is an infinite product of positive numbers), it must be that $L = \boxed{\textbf{(A) }\sqrt{10}}.$

~ Oxymoronic15

Solution 3

Move the first term inside the second radical. We get \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots = \sqrt[3]{10\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots.\] Do this for the third radical as well: \[\sqrt[3]{10\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots = \sqrt[3]{10\sqrt[3]{10}\sqrt[3]{\sqrt[3]{10}}} \cdots = \sqrt[3]{10\sqrt[3]{10\sqrt[3]{10\cdots}}}.\] It is clear what the pattern is. Setting the answer as $P,$ we have \[P = \sqrt[3]{10P},\] from which $P = \boxed{\sqrt{10}}.$

~kxiang

Solution 4

Set the product equal to P. We get \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots=P\] Since this is an infinite product, there may exist a clever manipulation where we set two different espressions involving $P$ equal, from which we could solve for a number. Since all terms are raised to the $\frac{1}{3}$ power, we can cube both sides of our equation. We would get \[10 \cdot \sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots=P^3\] From this, we know that $\frac{P^3}{10}=P$, and to this equation, the only solution is $P = \boxed{\sqrt{10}}$

~lmpofu

Video Solution 1 (HOW TO THINK CREATIVELY!!!)

https://youtu.be/_YDTIEuXTzY

~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

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