2000 AIME II Problems/Problem 1

From AoPSWiki

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

$\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}=\frac{\log_4{16}}{\log_4{2000^6}}+\frac{\log_5{125}}{\log_5{2000^6}}=\log_{2000^6}{2000}=\frac{1}{6}$

$1+6=\boxed{007}$

2000 AIME II (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Personal tools

Navigation

Search

Toolbox

2000 AIME II Problems/Problem 1

From AoPSWiki

Problem

Solution