Problem 1. Find the least positive integer $n$ with the following property: if $n$ distinct sums of the form $x_{p}+x_{q}+x_{r}, 1 \leq p<q<r \leq 5$, equal 0 , then $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0$.

Sava Grozdev, Svetlozar Doychev
1. If $x_{1}=2$ and $x_{2}=x_{3}=x_{3}=x_{5}=-1$, then all six sums of the form $x_{1}+x_{q}+x_{r}, 2 \leq q4$, there exists $x_{i}$ which occurs at least 5 times in these sums. We may assume that $i=1$.

The sums of the form $x_{1}+x_{q}+x_{r}, 2 \leq q<r \leq 5$, are 6 . Hence at most one of them is non-zero. So we may assume that

$$
\begin{aligned}
x_{1}+x_{2}+x_{4} & =x_{1}+x_{2}+x_{5}=x_{1}+x_{3}+x_{4} \\
& =x_{1}+x_{3}+x_{5}=x_{1}+x_{4}+x_{5}=0
\end{aligned}
$$

It follows that $x_{2}=x_{3}=x_{4}=x_{5}=-\frac{x_{1}}{2}$ So all the 6 sums $x_{1}+x_{q}+x_{r}$, $2 \leq q<r \leq 5$ equal 0 . Since the 7 -th sum which equals 0 has the form $x_{p}+x_{q}+x_{r}, 2 \leq p<q<r \leq 5$, we get that $3 x_{p}=0$, which implies that $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0$.