110. Find all ten-digit numbers such that the 1st digit of the number equals the number of zeros in its decimal representation, the 2nd digit - the number of ones, and so on up to the 10th digit, which equals the number of nines in the number's representation.
110. Let the desired number $X=\overline{a_{0} a_{1} a_{2} \ldots a_{9}}$ (where $a_{0}, a_{1}, \ldots, a_{9}$ are the digits of the number); in this case, $a_{0}$ is the number of zeros among the digits of $X$, $a_{1}$ is the number of ones, $a_{2}$ is the number of twos, and so on. Therefore, the sum of all the digits of $X$ is

$$
a_{0}+a_{1}+a_{2}+\ldots+a_{9}=a_{0} \cdot 0+a_{1} \cdot 1+a_{2} \cdot 2+\ldots+a_{9} \cdot 9
$$

from which we obtain

$$
a_{0}=a_{2}+2 a_{3}+3 a_{4}+4 a_{5}+5 a_{6}+6 a_{7}+7 a_{3}+8 a_{9}
$$

The equality $a_{0}=0$ is excluded by the conditions of the problem (otherwise the number $X$ would not be 10-digit; however, the condition $a_{0} \neq 0$ immediately follows from (*)); the equality $a_{0}=1$ leads to impossible values $a_{0}=a_{2}=1, a_{1}=8$ (since we should have 10 digits in total), all other digits are zeros; the equality $a_{0}=2$ is compatible only with (also impossible!) values $a_{0}=a_{2}=2, a_{1}=6$, all other digits are zeros, or $a_{0}=2, a_{3}=1, a_{1}=7$, all other digits are zeros. Let now $a_{0}=i>2$; we rewrite equality (*) as follows:

$$
a_{0}=i=a_{2}+2 a_{3}+\ldots+(i-1) a_{i}+\ldots+8 a_{9}
$$

(if $i=3$, then the terms $2 a_{3}$ and $(i-1) a_{i}$ coincide; if $i=9$, then $(i-1) a_{i}$ and $8 a_{9}$ coincide). In this case, $a_{i}$, the number of digits equal to $i$ in the number $X$, is non-zero (since $a_{0}=i$); on the other hand, (**) is impossible for $a_{i}>1$, so $a_{i}=1$. Therefore, (**) can be rewritten as:

$$
1=a_{2}+2 a_{3}+\ldots+(i-2) a_{i-1}+i a_{i+1}+\ldots+8 a_{9}, \quad(* * *)
$$

from which it immediately follows that $a_{2}=1$, and all digits of the number $X$ different from $a_{0}, a_{1}, a_{2}$ and $a_{i}$ are zero. But if $a_{2}=1$, then among the digits of the number $X$ there is a two, which, obviously, can only be the digit $a_{1}$. Thus, in the decimal representation of $X$, only the digits $a_{0}(=i), a_{1}=2, a_{2}=1$ and $a_{i}=1$ are non-zero, i.e., among the digits of $X$ there are $i$ zeros, 2 ones, 1 two, and 1 digit $i$, from which (since $X$ is 10-digit) $i=10-2-1-1=6$.

Thus, $X=6210001000$ (it is easy to check that this number indeed satisfies all the conditions of the problem).