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

Revision as of 19:51, 11 November 2022

The infinite product

$\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} ...$

evaluates to a real number. What is that number?

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

Revision as of 19:50, 11 November 2022 view source Phuang1024 (talk \| contribs) 12 edits Created solution.		Revision as of 19:51, 11 November 2022 view source Phuang1024 (talk \| contribs) 12 edits m →Solution: Add author. Newer edit →
Line 24:		Line 24:

	Thus, our answer is <math>10 ^ \frac{1}{2} = \boxed{(A) \sqrt{10}}</math>		Thus, our answer is <math>10 ^ \frac{1}{2} = \boxed{(A) \sqrt{10}}</math>

			- phuang1024