11. (40 points) Let's call the tail of a natural number any number that is obtained from it by discarding one or several of its first digits. For example, 234, 34, and 4 are tails of the number 1234. Does there exist a six-digit number without zeros in its decimal representation that is divisible by each of its tails?
Answer: Yes (721875 fits).

Solution. Suppose the required number exists. Let's write it as $A=$ $=\overline{a_{5} a_{4} \ldots a_{0}}$. Then $A$ is divisible by its five-digit tail, that is, by $\overline{a_{4} \ldots a_{0}}$. Therefore, $\overline{a_{4} \ldots a_{0}}$ divides the difference between $A$ and its tail, which is $a_{5} \cdot 10^{5}$. Thus, we need to check the five-digit divisors of $a_{5} \cdot 10^{5}$ without zeros in their decimal representation. Any divisor of the number $a_{5} \cdot 10^{5}$ is the product of $a_{5}$ and some powers of five and two. Since this product should not end in zero, it cannot contain both two and five simultaneously. If we take only twos, we get a number not exceeding $a_{5} \cdot 2^{5} \leqslant 9 \cdot 2^{5}=288$, which is not a five-digit number. Therefore, we need the number $a_{5} \cdot 5^{n}$, where $a_{5}$ is odd. Note that $n=5$, otherwise $a_{5} \cdot 5^{n} \leqslant 9 \cdot 625=5625$, and this number is also not five-digit. Finally, $a_{5}>3$, since the number $3 \cdot 5^{5}=9375$ is four-digit.

Considering the constraints, we get three possible options: $5 \cdot 5^{5}$, $7 \cdot 5^{5}, 9 \cdot 5^{5}$. The second one gives the number 721875, which fits:

$$
\begin{gathered}
721875 / 21875=33 ; 721875 / 1875=385 ; 721875 / 875=825 ; \\
721875 / 75=9625 ; 721875 / 5=144375
\end{gathered}
$$

It is not difficult to check that the other two options do not work. However, this is not necessary, as the answer has already been obtained.