2001 AMC 12 Problems/Problem 9
Contents
Problem
Let 
be a function satisfying 
for all positive real numbers 
and 
. If 
, what is the value of 
?
Solution 1
Letting 
and 
in the given equation, we get 
.
Solution 2a
no
Solution 2b
Having determined that 
for some constant 
, as in Solution 2a, an alternative way to finish the problem is to directly calculate: ![\[f(600) = \frac{k}{600} = \frac{5}{6}\cdot\frac{k}{500} = \frac{5}{6}f(500) = \frac{5}{6} \cdot 3 = \boxed{\textbf{(C) } \frac{5}{2}}.\]](http://latex.artofproblemsolving.com/5/9/3/593dfc969b7ed3e4dba78178fc096befb77e6e42.png)
Solution 3 (educated guessing)
Notice that 
and 
both have 
in common, so set 
and 
. Thus, we get ![\[f(500) = \frac{f(100)}{5} = 3 \iff f(100) = 15.\]](http://latex.artofproblemsolving.com/1/e/d/1edbaff1591b5b25abd93a1adb3c73d528d81d24.png)
Then, since we know ![\[f(600) = f(100\cdot6),\]](http://latex.artofproblemsolving.com/9/b/b/9bb85348718d9c01e2a2904985c49a0fadf6d20d.png)
we get ![\[f(600) = \frac{f(100)}{6} = \frac{15}{6} = \boxed{\textbf{(C) } \frac{5}{2}}.\]](http://latex.artofproblemsolving.com/3/7/c/37cad113f7337d7285bb81df05aecc93265a9f0f.png)
See Also
| 2001 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
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