Difference between revisions of "2022 AMC 12A Problems/Problem 8"
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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 | 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 | 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>. | Thus, our answer is <math>10 ^ \frac{1}{2} = \boxed{\textbf{(A) }\sqrt{10}}</math>. | ||
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We can write this infinite product as <math>L</math> (we know from the answer choices that the product must converge): | 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> | |
− | <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> | |
− | If we raise everything to the | + | 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> |
− | |||
− | <cmath>L^3 = 10 \, \cdot \, \sqrt[3]{10} \, \cdot \, \sqrt[3]{\sqrt[3]{10}} \cdots = 10L \implies L^3 - 10L = 0 \implies L \in \{0, \pm \sqrt{10}\}</cmath> | ||
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− | Since <math>L</math> is positive (it is an infinite product of positive numbers), it must be that <math>L = \boxed{\textbf{(A) }\sqrt{10}}</math> | ||
− | |||
− | |||
~ Oxymoronic15 | ~ Oxymoronic15 | ||
Line 52: | Line 40: | ||
~kxiang | ~kxiang | ||
− | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | + | ==Solution 4== |
+ | Set the product equal to P. We get | ||
+ | <cmath>\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots=P</cmath> Since this is an infinite product, there may exist a clever manipulation where we set two different espressions involving <math>P</math> equal, from which we could solve for a number. Since all terms are raised to the <math>\frac{1}{3}</math> power, we can cube both sides of our equation. We would get <cmath>10 \cdot \sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots=P^3</cmath> From this, we know that <math>\frac{P^3}{10}=P</math>, and to this equation, the only solution is <math>P = \boxed{\sqrt{10}}</math> | ||
+ | |||
+ | ~lmpofu | ||
+ | |||
+ | ==Video Solution 1 (HOW TO THINK CREATIVELY!!!)== | ||
https://youtu.be/_YDTIEuXTzY | https://youtu.be/_YDTIEuXTzY | ||
~Education, the Study of Everything | ~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 14:36, 5 October 2024
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
Solution 4
Set the product equal to P. We get Since this is an infinite product, there may exist a clever manipulation where we set two different espressions involving equal, from which we could solve for a number. Since all terms are raised to the power, we can cube both sides of our equation. We would get From this, we know that , and to this equation, the only solution is
~lmpofu
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.