3. $[x]$ represents the greatest integer not exceeding the real number $x$ (for example, $[\pi]=3,[-\pi]=-4,[-4]=-4$). Let $M=[x]+[2x]+[3x]$. Positive integers that cannot be expressed in the form of $M$ are called "invisible numbers". If the invisible numbers are arranged in ascending order, the 2009th invisible number is $\qquad$
3.6028 .

Let $n$ be a natural number.
When $n \leqslant x < n+\frac{1}{3}$,
$2 n \leqslant 2 x < 2 n+\frac{2}{3}$,
$3 n \leqslant 3 x < 3 n+1$.
Thus, $M=n+2 n+3 n=6 n$.
When $n+\frac{1}{3} \leqslant x < n+\frac{1}{2}$,
$2 n+\frac{2}{3} \leqslant 2 x < 2 n+1$,
$3 n+1 \leqslant 3 x < 3 n+\frac{3}{2}$.
Thus, $M=n+2 n+(3 n+1)=6 n+1$.
When $n+\frac{1}{2} \leqslant x < n+\frac{2}{3}$,
$2 n+1 \leqslant 2 x < 2 n+\frac{4}{3}$,
$3 n+\frac{3}{2} \leqslant 3 x < 3 n+2$.
Thus, $M=n+(2 n+1)+(3 n+1)=6 n+2$.
When $n+\frac{2}{3} \leqslant x < n+1$,
$2 n+\frac{4}{3} \leqslant 2 x < 2 n+2$,
$3 n+2 \leqslant 3 x < 3 n+3$.
Thus, $M=n+(2 n+1)+(3 n+2)=6 n+3$.
Therefore, the hidden numbers are $6 n+4$ or $6 n+5 (n \in \mathbf{N})$.
The 2009th hidden number is
$$
6 \times \frac{2009-1}{2}+4=6028
$$