Difference between revisions of "2020 AIME II Problems/Problem 3"
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https://youtu.be/lPr4fYEoXi0 ~ CNCM | https://youtu.be/lPr4fYEoXi0 ~ CNCM | ||
==Video Solution 2== | ==Video Solution 2== | ||
− | https://www.youtube.com/watch?v=x0QznvXcwHY | + | https://www.youtube.com/watch?v=x0QznvXcwHY?t=528 |
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==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=II|num-b=2|num-a=4}} | {{AIME box|year=2020|n=II|num-b=2|num-a=4}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:27, 8 June 2020
Contents
[hide]Problem
The value of that satisfies can be written as , where and are relatively prime positive integers. Find .
Solution
Let . Based on the equation, we get and . Expanding the second equation, we get . Substituting the first equation in, we get , so . Taking the 100th root, we get . Therefore, , so and the answer is . ~rayfish
Easiest Solution
Recall the identity (which is easily proven using exponents) Then this problem turns into Divide from both sides. And we are left with .Solving this simple equation we get ~mlgjeffdoge21
Video Solution
https://youtu.be/lPr4fYEoXi0 ~ CNCM
Video Solution 2
https://www.youtube.com/watch?v=x0QznvXcwHY?t=528
~IceMatrix
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.