# 6. Does there exist a natural number that is a multiple of 2015, the sum of whose digits is 2015?
Answer. Exists.

Solution. It is sufficient to provide one example of such a number. We will show a couple of ways to obtain such examples.

Example 1. Notice that $10075=2015 \cdot 5$, and the sum of the digits of the number 10075 is 13.

Then the number $\underbrace{1007510075 \ldots 10075}$ is divisible by 2015, and the sum of its digits is 155 times

$13 \cdot 155=2015$.

Example 2. The sum of the digits of the number 2015 is 8, and the sum of the digits of the number $4030=2015 \cdot 2$ is 7. Represent the number 2015 as the sum of two addends, one of which is divisible by seven, and the other by eight. For example, $2015=7+8 \cdot 251$. Then the number 403020152015...2015 is divisible by 2015, and the sum of its digits is 2015. 251 times

## Grading criteria.

- For a complete solution to the problem, it is sufficient to provide an example of a number and show that it meets the given requirements, -7 points.
- If a number is provided without explanations, but the jury can prove that it fits, -5 points.
- If the solution contains the idea of "constructing" the required number from numbers divisible by 2015, for example 403040302015...2015, but the number of numbers is calculated incorrectly, -3 points.
- Only the correct answer -0 points.