6. (36th Austrian Mathematical Olympiad) Let $f$ be a function defined on $\{0,1, \cdots, 2005\}$, taking values in the set of non-negative integers. For any variable $x$ in the domain, we have $f(2 x+1)=f(2 x)$, $f(3 x+1)=f(3 x)$, $f(5 x+1)=f(5 x)$. How many different function values can this function take at most?
6. For any $y \in\{0,1, \cdots, 2005\}$, if $y$ is divisible by $2,3,5$, then $f(y)=f(y+1)$ holds.

For a variable $y$, to make $f(y)$ and $f(y+1)$ not equal, then $(y, 30)=1$. Since $\varphi(30)=8,2005=30 \times 66+$ 25 , there are 
$$
66 \times 8+7=535 \text { (numbers) }
$$
in $\{0,1, \cdots, 2005\}$ that are coprime with 30.

It divides $\{0,1, \cdots, 2005\}$ into 536 parts, with the function value being equal in each part.
In conclusion, the function $f$ can take at most 536 different values.