Difference between revisions of "2000 AIME II Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | {{ | + | <math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}=\frac{\log_4{16}}{\log_4{2000^6}}+\frac{\log_5{125}}{\log_5{2000^6}}=\log_{2000^6}{2000}=\frac{1}{6}</math> |
| + | |||
| + | <math>1+6=\boxed{007}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2000|n=II|before=First Question|num-a=2}} | {{AIME box|year=2000|n=II|before=First Question|num-a=2}} | ||
Revision as of 12:45, 13 November 2007
Problem
The number
can be written as 
where 
and 
are relatively prime positive integers. Find 
.
Solution
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 | ||
