2022 AMC 12A Problems/Problem 8: Difference between revisions

Revision as of 21:30, 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 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

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/_YDTIEuXTzY

~Education, the Study of Everything

@@ Line 30: / Line 30: @@
 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>
-<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>
-If we raise everything to the <math>3^{rd}</math> power, we get:
+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^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>.
 ~ Oxymoronic15

2022 AMC 12A (Problems • Answer Key • Resources)
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