2022 AMC 12A Problems/Problem 8

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

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