Difference between revisions of "2000 AIME II Problems/Problem 1"

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== Solution ==
 
== Solution ==
{{solution}}
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=== Solution 1 ===
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<math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}</math>
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<math>=\frac{\log_4{16}}{\log_4{2000^6}}+\frac{\log_5{125}}{\log_5{2000^6}}</math>
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<math>=\frac{\log{16}}{\log{2000^6}}+\frac{\log{125}}{\log{2000^6}}</math>
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<math>=\frac{\log{2000}}{\log{2000^6}}</math>
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<math>=\frac{\log{2000}}{6\log{2000}}</math>
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<math>=\frac{1}{6}</math>
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Therefore, <math> m+n=1+6=\boxed{007}</math>
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=== Solution 2 ===
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Alternatively, we could've noted that, because <math>\frac 1{\log_a{b}} = \log_b{a}</math>
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<cmath>
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\begin{align*}
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\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} }\\
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&=2{\log_{2000^6}{4}} + 3{\log_{2000^6}{5}} \\
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&={\log_{2000^6}{4^2}} + {\log_{2000^6}{5^3}}\\
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&={\log_{2000^6}{4^2 \cdot 5^3}}\\
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&={\log_{2000^6}{2000}}\\
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&= {\frac{1}{6}}.\end{align*}</cmath>
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Therefore our answer is <math>1 + 6 = \boxed{7}</math>.
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== 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}}
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{{MAA Notice}}

Revision as of 18:41, 13 March 2015

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)
Preceded by
First Question
Followed by
Problem 2
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All AIME Problems and Solutions

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