Problem 4. If we increase the natural number $a$ by $10 \%$, we get the natural number $b$. Is it possible for the sum of the digits of $b$ to be $10 \%$ less than the sum of the digits of $a$?
Solution. We will find the number with the desired property, which is of the form

$$
a=\underbrace{88 \ldots 88}_{m} 0
$$

(written with $m$ eights and a zero). Let's increase $a$ by 10%:

$$
\begin{array}{rcccccc}
8 & 8 & \ldots & 8 & 8 & 8 & 0 \\
+ & 8 & 8 & \ldots & 8 & 8 & 8 \\
\hline 9 & 7 & 7 & \ldots & 7 & 6 & 8
\end{array}
$$

In the representation of the resulting number $b$, one digit 9, 6, and 8, and $m-2$ digits 7 are involved. The sum of the digits of $b$ is 10% less than the sum of the digits of $a$ precisely when the equation

$$
9+6+8+7(m-2)=90 \%(8 m)
$$

is satisfied. From this, we find $m=45$.

Note. The smallest number that satisfies the condition of the problem is 909999999990.