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Difference between revisions of "2000 AIME II Problems/Problem 1"
(→‎Solution)
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== Solution ==
 
== Solution ==
 +
=== Solution 1 ===
 
<math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}</math>
 
<math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}</math>
  
Line 18: Line 19:
  
 
Therefore, <math> m+n=1+6=\boxed{007}</math>
 
Therefore, <math> m+n=1+6=\boxed{007}</math>
 +
 +
=== Solution 2 ===
 +
 +
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>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>
 +
 +
<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}}

Revision as of 21:20, 30 December 2014

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 $\frac 1{\log_a{b}} = \log_b{a}$

\[\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} =\] \[2 \cdot \frac{1}{\log_4{2000^6}} + 3\cdot \frac {1}{\log_5{2000^6} }=\]

\[2{\log_{2000^6}{4}} + 3{\log_{2000^6}{5}} =\] \[{\log_{2000^6}{4^2}} + {\log_{2000^6}{5^3}} =\] 

\[{\log_{2000^6}{4^2 \cdot 5^3} =\] (Error compiling LaTeX. Unknown error_msg)
\[{\log_{2000^6}{2000} = {\frac{1}{6}}\] (Error compiling LaTeX. Unknown error_msg)

Therefore our answer is $1 + 6 = \boxed{7}$


2000 AIME II (Problems • Answer Key • Resources)
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First Question 		Followed by
Problem 2
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All AIME Problems and Solutions

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