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

Revision as of 20:32, 25 March 2011

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 $m + n$ .

$\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}$

$=\frac{\log_4{16}}{\log_4{2000^6}}+\frac{\log_5{125}}{\log_5{2000^6}}$

$=\frac{\log{16}}{\log{2000^6}}+\frac{\log{125}}{\log{2000^6}}$

$=\frac{\log{2000}}{\log{2000^6}}$

$=\frac{\log{2000}}{6\log{2000}}$

$=\frac{1}{6}$

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

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

@@ Line 7: / Line 7: @@
 <math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}</math>
-<math>\frac{\log_4{16}}{\log_4{2000^6}}+\frac{\log_5{125}}{\log_5{2000^6}}</math>
+<math>=\frac{\log_4{16}}{\log_4{2000^6}}+\frac{\log_5{125}}{\log_5{2000^6}}</math>
-<math>\frac{\log{16}}{\log{2000^6}}+\frac{\log{125}}{\log{2000^6}}</math>
+<math>=\frac{\log{16}}{\log{2000^6}}+\frac{\log{125}}{\log{2000^6}}</math>
-<math>\frac{\log{2000}}{\log{2000^6}}</math>
+<math>=\frac{\log{2000}}{\log{2000^6}}</math>
-<math>\frac{\log{2000}}{6\log{2000}}</math>
+<math>=\frac{\log{2000}}{6\log{2000}}</math>
-<math>\frac{1}{6}</math>
+<math>=\frac{1}{6}</math>
-<math>1+6=\boxed{007}</math>
+<math>=1+6=\boxed{007}</math>
 {{AIME box|year=2000|n=II|before=First Question|num-a=2}}