3. Find the smallest natural number $n$ such that the sum of the digits of each of the numbers $n$ and $n+1$ is divisible by 17.
Answer: 8899.

Solution. If $n$ does not end in 9, then the sums of the digits of the numbers $n$ and $n+1$ differ by 1 and cannot both be divisible by 17. Let $n=\overline{m 99 \ldots 9}$, where the end consists of $k$ nines, and the number $m$ has a sum of digits $s$ and does not end in 9. Then the sum of the digits of the number $n$ is $s+9k$, and the sum of the digits of the number $n+1$ is clearly $s+1$. According to the condition, these two sums are divisible by 17, in particular, the number $s+1$ is a multiple of 17 $\Longrightarrow s \geqslant 16$. The smallest number $m$ that does not end in 9 and has a sum of digits of 16 is 88.

Next, the number $s+9k=(s+1)+(9k-1)$ is divisible by 17. Since the first term is divisible by 17, the number $9k-1$ is also a multiple of 17. The smallest value of $k$ that satisfies this condition is 2, that is, $n=8899$. The sums of the digits of each of the numbers $n$ and $n+1$ are divisible by 17.

Criteria. It is proven that the sought number ends in the digit $9-1$ point. Correct answer -2 points. It is proven that the sum of the digits, different from 9, is not less than 16 - another 2 points. It is proven that there are exactly two nines - another 2 points. Full solution -7 points.