19. If a positive integer $N$ can be expressed as $\lfloor x\rfloor+\lfloor 2 x\rfloor+\lfloor 3 x\rfloor$ for some real numbers $x$, then we say that $N$ is "visible"; otherwise, we say that $N$ is "invisible". For example, 8 is visible since $8=\lfloor 1.5\rfloor+\lfloor 2(1.5)\rfloor+\lfloor 3(1.5)\rfloor$, whereas 10 is invisible. If we arrange all the "invisible" positive integers in increasing order, find the $2011^{\text {th }}$ "invisible" integer.
19. Answer. 6034
Solution. Let $n$ be a positive integer.
If $n \leq x<n+\frac{1}{3}$, then $2 n \leq 2 x<2 n+\frac{2}{3}$ and $3 n \leq 3 x<3 n+1$, giving
$$
N=\lfloor x\rfloor+\lfloor 2 x\rfloor+\lfloor 3 x\rfloor=n+2 n+3 n=6 n \text {. }
$$

If $n+\frac{1}{3} \leq x<n+\frac{1}{2}$, then $2 n+\frac{2}{3} \leq 2 x<2 n+1$ and $3 n+1 \leq 3 x<3 n+\frac{3}{2}$, giving
$$
N=\lfloor x\rfloor+\lfloor 2 x\rfloor+\lfloor 3 x\rfloor=n+2 n+3 n+1=6 n+1 .
$$

If $n+\frac{1}{2} \leq x<n+\frac{2}{3}$, then $2 n+1 \leq 2 x<2 n+\frac{4}{3}$ and $3 n+\frac{3}{2} \leq 3 x<3 n+\frac{4}{3}$, giving
$$
N=\lfloor x\rfloor+\lfloor 2 x\rfloor+\lfloor 3 x\rfloor=n+2 n+1+3 n+1=6 n+2 .
$$

If $n+\frac{2}{3} \leq x<n+1$, then $2 n+\frac{4}{3} \leq 2 x<2 n+2$ and $3 n+2 \leq 3 x<3 n+3$, giving
$$
N=\lfloor x\rfloor+\lfloor 2 x\rfloor+\lfloor 3 x\rfloor=n+2 n+1+3 n+2=6 n+3 .
$$

Thus, "invisible" numbers must be of the form $6 n+4$ and $6 n+5$. The $2011^{\text {th }}$ "invisible" integer is $4+6 \times \frac{2011-1}{2}=6034$.