2000 AIME II Problems/Problem 1
Contents
Problem
The number
can be written as 
where 
and 
are relatively prime positive integers. Find 
.
Solution
Solution 1
Therefore, 
Solution 2
Alternatively, we could've noted that, because 
![\[\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} =\]](http://latex.artofproblemsolving.com/b/1/5/b156da5ef4a99dc1a0230a914fc839d816c2fca0.png)
![\[2 \cdot \frac{1}{\log_4{2000^6}} + 3\cdot \frac {1}{\log_5{2000^6} }=\]](http://latex.artofproblemsolving.com/d/2/9/d297d8832022780d8c27ab3aa2dc189dd4c56de4.png)
![\[2{\log_{2000^6}{4}} + 3{\log_{2000^6}{5}} =\]](http://latex.artofproblemsolving.com/3/8/d/38d2287f4531eb791ae6cfefdcaa7fccb71c0916.png)
![\[{\log_{2000^6}{4^2}} + {\log_{2000^6}{5^3}} =\]](http://latex.artofproblemsolving.com/9/5/4/954122277dd7550b40248ce493a9731551696b50.png)
\[{\log_{2000^6}{4^2 \cdot 5^3} =\] (Error compiling LaTeX. Unknown error_msg)
\[{\log_{2000^6}{2000} = {\frac{1}{6}}\] (Error compiling LaTeX. Unknown error_msg)
Therefore our answer is 
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