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 $1 + 6 = \boxed{007}$ .

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

ᴠɪᴅᴇᴏ ѕᴏʟᴜᴛɪᴏɴ ʙʏ ᴘɪ ᴀᴄᴀᴅᴇᴍʏ

https://youtu.be/ucn9yfcu1QY?si=r3ebuzJNd2uAq0kV

~ Pi Academy

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.

Page

Toolbox

Search