10.3. Find the smallest natural number divisible by 99, all digits of which are even.
Answer: 228888.

Solution. Let the sum of the digits of the desired number $X$, located in even-numbered positions (tens, thousands, etc.), be denoted by $A$, and the sum of the digits in odd-numbered positions (units, hundreds, etc.) be denoted by $B$. According to the divisibility rules for 9 and 11, $A+B$ is divisible by 9, and $A-B$ is divisible by 11. Given the evenness of the digits of the number, $A$ and $B$ are even, so $A+B$ is also divisible by 18, and $A-B$ is divisible by 22. If $A-B=0$, then $A$ and $B$ are equal and divisible by 9, and considering their evenness, $A$ and $B$ are divisible by 18. From this, it is easy to see that in this case, the desired number is not less than 228888, and the digits of the number 228888 are even and it is divisible by 9. Therefore, if there is such a number less than 228888, then for it $A \leq 18, B \leq 24$. Thus, if $A-B$ is not equal to 0, then $A-B=-22$, and either $A=0, B=22$, or $A=2, B=24$. In both cases, $A+B$ is not divisible by 9. Therefore, $X=228888$.