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

(Solution)
m (Solution 2)
 
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Alternatively, we could've noted that, because <math>\frac 1{\log_a{b}} = \log_b{a}</math>
 
Alternatively, we could've noted that, because <math>\frac 1{\log_a{b}} = \log_b{a}</math>
  
<cmath>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} = </cmath> <cmath>2 \cdot \frac{1}{\log_4{2000^6}} + 3\cdot \frac {1}{\log_5{2000^6} }= </cmath>
+
<cmath>
 +
\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*}</cmath>
  
<cmath>2{\log_{2000^6}{4}} + 3{\log_{2000^6}{5}} = </cmath> <cmath>{\log_{2000^6}{4^2}} + {\log_{2000^6}{5^3}} = </cmath> <cmath>{\log_{2000^6}{4^2 \cdot 5^3} = </cmath>
+
Therefore our answer is <math>1 + 6 = \boxed{7}</math>.
 
 
<cmath>{\log_{2000^6}{2000} = {\frac{1}{6}}</cmath>
 
 
 
Therefore our answer is <math>1 + 6 = \boxed{7}</math>
 
  
  
 
{{AIME box|year=2000|n=II|before=First Question|num-a=2}}
 
{{AIME box|year=2000|n=II|before=First Question|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest 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)
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First Question
Followed by
Problem 2
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