2022 AMC 12A Problems/Problem 8

Problem

The infinite product $\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \ldots$ 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

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

$10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} ...$

which is

$10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} ...}$ .

Using the formula for an infinite geometric series $S = \frac{a}{1-r}$ , we get

$\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} ... = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}$

Thus, our answer is $10 ^ \frac{1}{2} = \boxed{(A) \sqrt{10}}$

- phuang1024

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2022 AMC 12A Problems/Problem 8

Problem

Solution

See Also