# Funky Functional Equation

**Algebra**Level 5

Let \(f:\mathbb{R^+} \rightarrow \mathbb{R^+}\) be a function satisfying

\(f(1)=2\)

\(f(2)=1\)

\(f(3n)=3f(n)\)

\(f(3n+1)=3f(n)+2\)

\(f(3n+2)=3f(n)+1.\)

Find the number of integers \(n \leq 2006\) such that \(f(n)=2n.\)

See Part 2 if you enjoyed this problem! :D