## Task B-2.4.

In a mathematics competition, $n$ problems are solved, $10<n<20$. For each correctly solved problem, a contestant receives 5 points, and for each unsolved or incorrectly solved problem, they lose 3 points. If Sebastijan had a total of 0 points, how many problems did he solve correctly? How many problems were there in total?
## First solution.

Let $x$ be the number of tasks solved correctly, and $y$ be the number of tasks solved incorrectly or not solved.

Then $10 < x + y < 20, x, y \in \mathbb{N}$.

The total number of points is $5x - 3y$.

Since Sebastijan has 0 points, it follows that

$$
5x - 3y = 0, \text{ or } x = \frac{3y}{5}
$$

Since $x$ must be a natural number, it follows that $y$ must be divisible by 5.

Substitute $x = \frac{3y}{5}$ into $10 < x + y < 20$.

Then $10 < \frac{8y}{5} < 20$, or $6.25 < y < 12.5$. 1 point

Since $y$ is a natural number divisible by 5, the only possibility is $y = 10$. 1 point

Then $x = \frac{3y}{5} = 6$, and the total number of tasks is 16.