# 2000 AIME II Problems/Problem 1

## 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

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

Therefore, $m+n=1+6=\boxed{007}$

### Solution 2

Alternatively, we could've noted that, because $\frac 1{\log_a{b}} = \log_b{a}$ \begin{align*} \frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} &= 2 \cdot \frac{1}{\log_4{2000^6}} + 3\cdot \frac {1}{\log_5{2000^6} }\\ &=2{\log_{2000^6}{4}} + 3{\log_{2000^6}{5}} \\ &={\log_{2000^6}{4^2}} + {\log_{2000^6}{5^3}}\\ &={\log_{2000^6}{4^2 \cdot 5^3}}\\ &={\log_{2000^6}{2000}}\\ &= {\frac{1}{6}}.\end{align*}

Therefore our answer is $1 + 6 = \boxed{007}$.

## Solution 3

We know that $2 = \log_4{16}$ and $3 = \log_5{125}$, and by base of change formula, $\log_a{b} = \frac{\log_c{b}}{\log_c{a}}$. Lastly, notice $\log a + \log b = \log ab$ for all bases. \begin{align*} \frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} = \log_{2000^6}{16} + \log_{2000^6}{125} = \log_{2000^6}{2000} = \frac16 \implies \boxed{007} \end{align*} $\bold{Solution}$ $\bold{written}$ $\bold{by}$

~ $\bold{PaperMath}$

 2000 AIME II (Problems • Answer Key • Resources) Preceded byFirst Question Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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