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 $1 + 6 = \boxed{7}$.


2000 AIME II (ProblemsAnswer KeyResources)
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First Question
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Problem 2
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