5. How many natural numbers from 1 to 2015 inclusive have a sum of digits that is a multiple of 5?
Answer: 402.

Solution: Note that among ten numbers of the form $\overline{a 0}, \ldots, \overline{a 9}$, exactly two numbers have a sum of digits that is a multiple of five. Thus, among the numbers from 10 to 2009, there are exactly 200 such tens, and therefore, 400 such numbers. Considering also the numbers 5 and 2012, we get a total of 402 such numbers.