Difference between revisions of "2020 AIME II Problems/Problem 3"
Line 13: | Line 13: | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/lPr4fYEoXi0 ~ CNCM | https://youtu.be/lPr4fYEoXi0 ~ CNCM | ||
+ | ==Video Solution 2== | ||
+ | https://www.youtube.com/watch?v=x0QznvXcwHY | ||
==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 23:51, 7 June 2020
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
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.