Difference between revisions of "2022 AMC 12A Problems/Problem 8"
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The infinite product | The infinite product | ||
+ | <cmath>\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots</cmath> | ||
+ | evaluates to a real number. What is that number? | ||
− | <math>\sqrt | + | <math>\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}</math> |
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− | |||
− | ==Solution== | + | ==Solution 1== |
We can write <math>\sqrt[3]{10}</math> as <math>10 ^ \frac{1}{3}</math>. Similarly, <math>\sqrt[3]{\sqrt[3]{10}} = (10 ^ \frac{1}{3}) ^ \frac{1}{3} = 10 ^ \frac{1}{3^2}</math>. | We can write <math>\sqrt[3]{10}</math> as <math>10 ^ \frac{1}{3}</math>. Similarly, <math>\sqrt[3]{\sqrt[3]{10}} = (10 ^ \frac{1}{3}) ^ \frac{1}{3} = 10 ^ \frac{1}{3^2}</math>. | ||
By continuing this, we get the form | By continuing this, we get the form | ||
+ | <cmath>10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \cdots,</cmath> | ||
+ | which is | ||
+ | <cmath>10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots}.</cmath> | ||
+ | Using the formula for an infinite geometric series <math>S = \frac{a}{1-r}</math>, we get | ||
+ | <cmath>\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}.</cmath> | ||
+ | Thus, our answer is <math>10 ^ \frac{1}{2} = \boxed{\textbf{(A) }\sqrt{10}}</math>. | ||
− | + | - phuang1024 | |
− | which | + | ==Solution 2== |
+ | |||
+ | We can write this infinite product as <math>L</math> (we know from the answer choices that the product must converge): | ||
+ | <cmath>L = \sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots.</cmath> | ||
+ | If we raise everything to the third power, we get: | ||
+ | <cmath>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\}.</cmath> | ||
+ | Since <math>L</math> is positive (as it is an infinite product of positive numbers), it must be that <math>L = \boxed{\textbf{(A) }\sqrt{10}}.</math> | ||
+ | |||
+ | ~ Oxymoronic15 | ||
+ | |||
+ | ==Solution 3== | ||
+ | Move the first term inside the second radical. We get | ||
+ | <cmath>\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.</cmath> | ||
+ | Do this for the third radical as well: | ||
+ | <cmath>\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}}}.</cmath> | ||
+ | It is clear what the pattern is. Setting the answer as <math>P,</math> we have <cmath>P = \sqrt[3]{10P},</cmath> from which <math>P = \boxed{\sqrt{10}}.</math> | ||
+ | |||
+ | ~kxiang | ||
− | + | ==Video Solution 1 (HOW TO THINK CREATIVELY!!!)== | |
+ | https://youtu.be/_YDTIEuXTzY | ||
− | + | ~Education, the Study of Everything | |
− | + | ==Video Solution 2 (Smart and Fun!!!)== | |
+ | https://youtu.be/7yAh4MtJ8a8?si=OJHbJh4_xMjBc9OY&t=1397 | ||
− | + | ~Math-X | |
− | |||
== See Also == | == See Also == |
Latest revision as of 12:34, 25 October 2023
Contents
Problem
The infinite product
evaluates to a real number. What is that number?
Solution 1
We can write as
. Similarly,
.
By continuing this, we get the form
which is
Using the formula for an infinite geometric series
, we get
Thus, our answer is
.
- phuang1024
Solution 2
We can write this infinite product as (we know from the answer choices that the product must converge):
If we raise everything to the third power, we get:
Since
is positive (as it is an infinite product of positive numbers), it must be that
~ Oxymoronic15
Solution 3
Move the first term inside the second radical. We get
Do this for the third radical as well:
It is clear what the pattern is. Setting the answer as
we have
from which
~kxiang
Video Solution 1 (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
Video Solution 2 (Smart and Fun!!!)
https://youtu.be/7yAh4MtJ8a8?si=OJHbJh4_xMjBc9OY&t=1397
~Math-X
See Also
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.