## Task B-1.6.

A student has passed 31 exams over the course of a five-year study. Each year, he took more exams than the previous year, and in the fifth year, he took three times as many exams as in the first year. How many exams did the student take in the fourth year?
## Solution.

Let's denote the number of exams the student passed in the $i$-th year by $x_{i}, i \in\{1,2,3,4,5\}$.

Then $x_{5}=3 x_{1}, x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$ and $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=31$.

1 point

If $x_{1}=1$, then $x_{5}=3$. This is impossible because there are no 3 different natural numbers between 1 and 3.

If $x_{1}=2$, then $x_{5}=6$, and the number of exams in the remaining years must be 3, 4, and 5. However, the sum $2+3+4+5+6$ is not 31.

If $x_{1}=3$, then $x_{5}=9$, so $x_{2}, x_{3}, x_{4}$ are some of the numbers $4,5,6,7,8$. Since $3+9=12$, it must be that $x_{2}+x_{3}+x_{4}=19$, which can be obtained for $4+7+8$ or $5+6+8$.

For the case when $x_{1} \geqslant 4$, there is no solution, because $x_{5}=3 \cdot x_{1} \geqslant 12$, and the sum $x_{2}+x_{3}+x_{4}=$ $31-x_{1}-x_{5} \leqslant 15$. On the other hand, the smallest values that the numbers $x_{2}, x_{3}, x_{4}$ can take are $5,6,7,8,9,10,11, \ldots$, and their sum is at least 18, which is greater than 15.

The student passed 8 exams in the fourth year.