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Difference between revisions of "2022 AMC 12A Problems/Problem 8"

m (→‎Solution 1)
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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}}} \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?
  
Line 13: Line 13:
 
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} \cdot ...</math>
+
<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>.
+
<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} + ... = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}</math>
+
<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}}} \ldots</cmath>
+
<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}} \ldots = 10L \implies L^3 - 10L = 0 \implies L \in \{0, \pm \sqrt{10}\}</cmath>
+
<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>.
Line 45: Line 45:
 
==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}}} \ldots = \sqrt[3]{10\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 = \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}}} \ldots = \sqrt[3]{10\sqrt[3]{10}\sqrt[3]{\sqrt[3]{10}}} \ldots = \sqrt[3]{10\sqrt[3]{10\sqrt[3]{10\ldots}}}</cmath>
+
<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>
 
−
<cmath>P = \boxed{\sqrt{10}}</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 \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\] 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 $3^{rd}$ power, we get:

\[L^3 = 10 \, \cdot \, \sqrt[3]{10} \, \cdot \, \sqrt[3]{\sqrt[3]{10}} \cdots = 10L \implies L^3 - 10L = 0 \implies L \in \{0, \pm \sqrt{10}\}\]

Since $L$ is positive (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 \[P = \sqrt[3]{10P} = \boxed{\sqrt{10}}.\] ~kxiang

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/_YDTIEuXTzY

~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

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