Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 3)
Line 44: Line 44:
  
 
~ <math>\bold{PaperMath}</math>
 
~ <math>\bold{PaperMath}</math>
 +
 +
==Solution 4==
 +
<cmath>\frac{2}{\log_4 2000^6} + \frac{3}{\log_5 2000^6}</cmath> <cmath>= \frac{1}{3\log_4 2000} + \frac{1}{2\log_5 2000}</cmath> <cmath>= \frac{1}{3} \log_{2000} 4 + \frac{1}{2} \log_{2000} 5</cmath> <cmath>= \log_{2000} (\sqrt[3]{4} \cdot \sqrt{5}) = x</cmath> <cmath>\implies 2^{4x} \cdot 5^{3x} = 2^{\frac{2}{3}} \cdot 5^{\frac{1}{2}}</cmath> <cmath>\implies 4x + (3\log_2 5)x = \frac{2}{3}+\frac{1}{2} \log_2 5</cmath> <cmath>\implies x = \frac{\frac{2}{3} + \frac{1}{2} \log_2 5}{4 + 3\log_2 5}</cmath> <cmath>\implies 6x = \frac{4 + 3\log_2 5}{4 + 3\log_2 5}</cmath> <cmath>\implies x = \frac{1}{6}</cmath> <cmath>\implies m + n = \boxed{007}</cmath>
 +
 +
~ cxsmi
  
 
{{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 14:56, 11 February 2024

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*}

$\bold{Solution}$ $\bold{written}$ $\bold{by}$

~ $\bold{PaperMath}$

Solution 4

\[\frac{2}{\log_4 2000^6} + \frac{3}{\log_5 2000^6}\] \[= \frac{1}{3\log_4 2000} + \frac{1}{2\log_5 2000}\] \[= \frac{1}{3} \log_{2000} 4 + \frac{1}{2} \log_{2000} 5\] \[= \log_{2000} (\sqrt[3]{4} \cdot \sqrt{5}) = x\] \[\implies 2^{4x} \cdot 5^{3x} = 2^{\frac{2}{3}} \cdot 5^{\frac{1}{2}}\] \[\implies 4x + (3\log_2 5)x = \frac{2}{3}+\frac{1}{2} \log_2 5\] \[\implies x = \frac{\frac{2}{3} + \frac{1}{2} \log_2 5}{4 + 3\log_2 5}\] \[\implies 6x = \frac{4 + 3\log_2 5}{4 + 3\log_2 5}\] \[\implies x = \frac{1}{6}\] \[\implies m + n = \boxed{007}\]

~ cxsmi

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_1&oldid=215244"