Difference between revisions of "2022 AIME II Problems/Problem 4"
(→Solution 2) |
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Line 22: | Line 22: | ||
We have | We have | ||
+ | <cmath> | ||
\begin{align*} | \begin{align*} | ||
\log_{20x} (22x) | \log_{20x} (22x) | ||
Line 27: | Line 28: | ||
& = \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} . | & = \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} . | ||
\end{align*} | \end{align*} | ||
+ | </cmath> | ||
We have | We have | ||
+ | <cmath> | ||
\begin{align*} | \begin{align*} | ||
\log_{2x} (202x) | \log_{2x} (202x) | ||
Line 34: | Line 37: | ||
& = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} . | & = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} . | ||
\end{align*} | \end{align*} | ||
+ | </cmath> | ||
Because <math>\log_{20x} (22x)=\log_{2x} (202x)</math>, we get | Because <math>\log_{20x} (22x)=\log_{2x} (202x)</math>, we get | ||
+ | <cmath> | ||
\[ | \[ | ||
\frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} | \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} | ||
= \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} . | = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} . | ||
\] | \] | ||
+ | </cmath> | ||
We denote this common value as <math>\lambda</math>. | We denote this common value as <math>\lambda</math>. |
Revision as of 14:31, 18 February 2022
Contents
Problem
There is a positive real number not equal to either
or
such that
The value
can be written as
, where
and
are relatively prime positive integers. Find
.
Solution 1
We could assume a variable which equals to both
and
.
So that
and
Express as:
Substitute to
:
Thus, , where
and
.
Therefore, .
~DSAERF-CALMIT (https://binaryphi.site)
Solution 2
We have
We have
Because , we get
We denote this common value as .
By solving the equality , we get
.
By solving the equality , we get
.
By equating these two equations, we get
Therefore,
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.