Difference between revisions of "2022 AMC 12A Problems/Problem 8"

Revision as of 16:42, 15 October 2023

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 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

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

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):

$L = \sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \ldots$

If we raise everything to the $3^{rd}$ power, we get:

$L^3 = 10 \, \cdot \, \sqrt[3]{10} \, \cdot \, \sqrt[3]{\sqrt[3]{10}} \ldots = 10L \implies L^3 - 10L = 0 \implies L \in \{0, \pm \sqrt{10}\}$

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 $\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \ldots = \sqrt[3]{10\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \ldots$

Do this for the third radical as well.

$\sqrt[3]{10\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \ldots = \sqrt[3]{10\sqrt[3]{10}\sqrt[3]{\sqrt[3]{10}}} \ldots = \sqrt[3]{10\sqrt[3]{10\sqrt[3]{10\ldots}}}$

It is clear what the pattern is. Setting the answer as $P,$ we have

$P = \sqrt[3]{10P}$ $P = \boxed{\sqrt{10}}$

~kxiang

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/_YDTIEuXTzY

~Education, the Study of Everything

@@ Line 13: / Line 13: @@
 By continuing this, we get the form
-<math>10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} ...</math>
+<math>10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \cdot ...</math>
 which is
-<math>10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} ...}</math>.
+<math>10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + ...}</math>.
 Using the formula for an infinite geometric series <math>S = \frac{a}{1-r}</math>, we get
-<math>\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} ... = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}</math>
+<math>\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + ... = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}</math>
 Thus, our answer is <math>10 ^ \frac{1}{2} = \boxed{\textbf{(A) }\sqrt{10}}</math>.

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

Page

Toolbox

Search