Difference between revisions of "2000 AIME II Problems/Problem 1"
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== Problem == | == Problem == | ||
− | + | The number | |
+ | <center><math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}</math></center> | ||
+ | can be written as <math>\frac mn</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | ||
== Solution == | == Solution == |
Revision as of 19:21, 11 November 2007
Problem
The number
![$\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}$](http://latex.artofproblemsolving.com/1/c/a/1ca16383129aebb061bb0e345327dff7fbc71548.png)
can be written as where
and
are relatively prime positive integers. Find
.
Solution
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See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |