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2000 AIME II Problems/Problem 1

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Problem

The number

\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}

can be written as \frac mn where m and n are relatively prime positive integers. Find m + n.

Solution

\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}}

\frac{\log{16}}{\log{2000^6}}+\frac{\log{125}}{\log{2000^6}}

\frac{\log{2000}}{\log{2000^6}}

\frac{\log{2000}}{6\log{2000}}

\frac{1}{6}

1+6=\boxed{007}

2000 AIME II (Problems • Resources)
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First Question 		Followed by
Problem 2
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