# 2022 AMC 12A Problems/Problem 8

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

The infinite product $$\sqrt{10} \cdot \sqrt{\sqrt{10}} \cdot \sqrt{\sqrt{\sqrt{10}}} \ldots$$ evaluates to a real number. What is that number? $\textbf{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt{100}\qquad\textbf{(C) }\sqrt{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt{10}$

## Solution 1

We can write $\sqrt{10}$ as $10 ^ \frac{1}{3}$. Similarly, $\sqrt{\sqrt{10}} = (10 ^ \frac{1}{3}) ^ \frac{1}{3} = 10 ^ \frac{1}{3^2}$.

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 $10 ^ \frac{1}{2} = \boxed{\textbf{(A) }\sqrt{10}}$.

- phuang1024

## Solution 2

We can write this infinite product as $L$ (we know from the answer choices that the product must converge): $$L = \sqrt{10} \cdot \sqrt{\sqrt{10}} \cdot \sqrt{\sqrt{\sqrt{10}}} \ldots$$

If we raise everything to the $3^{rd}$ power, we get: $$L^3 = 10 \, \cdot \, \sqrt{10} \, \cdot \, \sqrt{\sqrt{10}} \ldots = 10L \implies L^3 - 10L = 0 \implies L \in \{0, \pm \sqrt{10}\}$$

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 $$\sqrt{10} \cdot \sqrt{\sqrt{10}} \cdot \sqrt{\sqrt{\sqrt{10}}} \ldots = \sqrt{10\sqrt{10}} \cdot \sqrt{\sqrt{\sqrt{10}}} \ldots$$

Do this for the third radical as well. $$\sqrt{10\sqrt{10}} \cdot \sqrt{\sqrt{\sqrt{10}}} \ldots = \sqrt{10\sqrt{10}\sqrt{\sqrt{10}}} \ldots = \sqrt{10\sqrt{10\sqrt{10\ldots}}}$$

It is clear what the pattern is. Setting the answer as $P,$ we have $$P = \sqrt{10P}$$ $$P = \boxed{\sqrt{10}}$$

~kxiang

## Video Solution (HOW TO THINK CREATIVELY!!!)

~Education, the Study of Everything

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