10. All integers from 1 to 222222222 are written in sequence. How many times does the digit 0 appear in the recording of these numbers?

## 2. Factorization of numbers into products and their decomposition into sums

In solving some of the subsequent problems, the following notations prove useful.

The symbol $[x]$ (read as "the integer part of $x$") denotes the greatest integer not exceeding $x$. For example,

$\left[\frac{3}{2}\right]=1, \quad[10.85]=10, \quad[5]=5, \quad[-8.2]=-9$, etc.

(however, we will only need the symbol $[x]$ for positive numbers $x$).

The symbol $(x)$ (read as "the nearest integer to $x$") denotes the integer closest to the number $x$. For example,

$$
(5.4)=5, \quad(8.73)=9, \quad(6)=6, \quad(-2.8)=-3
$$

and so on.

Clearly, $(x)$ will be equal to $[x]$ or $[x]+1$ depending on whether the difference $x - [x]$ is less than or greater than half. In the case where $x - [x] = \frac{1}{2}$, $(x)$ can be taken as either $[x]$ or $[x]+1$; in this case, it is agreed to consider $(x) = [x]$ (however, we will use the symbol $(x)$ only in cases where $x - [x] \neq \frac{1}{2}$, so there will be no ambiguity as to whether to consider $(x)$ equal to $[x]$ or $[x]+1$).

In all subsequent problems where the number $n$ is mentioned in the condition, this number is considered to be a positive integer.
10. Obviously, among the first 222222222 integers, there will be 22222222 numbers ending in zero (numbers 10, $20,30, \ldots, 222222220$). Further, among them, there will be 2222222 numbers ending in the digits 00 (numbers $100,200,300, \ldots, 222222200$); 2222222 numbers ending in the digits 01 (numbers 101, 201, 301, ..., 222222201); 2222222 numbers ending in the digits 02; 2222222 numbers ending in the digits 03, and so on, finally, 2222222 numbers ending in the digits 09. Thus, in the sequence of all numbers from 1 to 222222222, the digit 0 will stand in the second-to-last place $2222222+2222222+\ldots+2222222=$ $=10 \cdot 2222222=22222220$ times.

Similarly, among the first 222222222 numbers, there will be 222222 numbers ending in the digits 000; 222222 numbers ending in the digits 001, and so on, finally, 222222 numbers ending in the digits 099. Thus, the digit 0 will stand in the third-to-last place $100 \cdot 222222=22222200$ times. Further, it is shown that the digit 0 will stand in the fourth-to-last place $1000 \cdot 22222=22222000$ times; in the fifth-to-last place $10000 \cdot 2222=22220000$ times; in the sixth-to-last place $100000 \cdot 222=22200000$ times; in the seventh-to-last place $1000000 \cdot 22=22000000$ times; finally, in the eighth-to-last place $10000000 \cdot 2=20000000$ times.

Thus, the digit 0 appears a total of

$22222222+22222220+22222200+22222000+$

$+22220000+22200000+22000000+20000000=$

$=175308642$

times.