8.1. Find the smallest natural number divisible by 5, with the sum of its digits being 100. Justify your answer.
Answer: 599999999995 (between two fives there are 10 nines).

Solution. Due to divisibility by 5, the last digit of the number $N$ can be either 5 or 0. If the last digit is 0, without changing the sum of the digits, we can replace 0 with 5 and subtract one from each of the five non-zero digits in other positions. Then we get a smaller number that satisfies the conditions of the problem. Next, if any digit of $N$ between the first and last is not a nine, then by increasing it by one and decreasing the digit in the higher position by one, we again get a smaller number. Thus, all digits of $N$ except the first and last are nines (say, $n$ of them), and the last one is a five. The first digit $x$ is found from the condition for the sum of the digits: $x+9n+5=100$, i.e., $9n=95-x$. Obviously, the digit $x$ for the divisibility of the right side by 9 is uniquely determined: $x=5$. Hence the answer.