5. Does there exist a natural number that, when divided by the sum of its digits, gives both a quotient and a remainder of 2014? If there is more than one such number, write their sum as the answer. If no such numbers exist, write 0. (12 points)

#
# Solution:

Suppose there exists a natural number $\boldsymbol{n}$ with the sum of its digits $\boldsymbol{s}$, such that $n=2014 s+2014$, from which we get $n-s=2013 s+2014$. By the divisibility rule, $n-s$ is divisible by 3. However, the number $2013 s+2014$ is not divisible by 3, since the number $2013s$ is a multiple of 3, while the number 2014 is not. This is a contradiction.

Answer: 0.

| Points | Criteria for evaluating the performance of task № $\mathbf{5}$ |
| :---: | :--- |
| $\mathbf{1 2}$ | A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are performed correctly. The correct answer is obtained. |
| $\mathbf{1 0}$ | A correct sequence of all steps of the solution is provided. There are gaps in the justification of divisibility by 3 or a computational error or typo that does not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. |
| $\mathbf{5}$ | The problem is not solved, but its solution is significantly advanced, i.e.:  - a significant part of the solution is performed correctly, possibly inaccurately (for example, the equation for the desired number is correctly written);  - the other part is either not performed or performed incorrectly, possibly even with logical errors (for example, the desired number is found or its absence is incorrectly justified).  The solution may be incomplete. |
| $\mathbf{0}$ | All cases of the solution that do not correspond to the above criteria for scoring 5, 10, and 12 points. |