Could you help me with the following task?
Prove that for four numbers form a set $S = \{ \alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5, \alpha_6 \}$ of distinct integers it is possible to choose four numbers of $S$ such that $1\cdot \alpha_1 + 2\cdot \alpha_2 + 3\cdot \alpha_3 + 4\cdot \alpha_4 = 5\cdot \beta$, where $\beta \in \mathbb{Z}$
At first, I have tried to edit the expression:
$$\begin{matrix} 1\cdot \alpha_1 + 2\cdot \alpha_2 + 3\cdot \alpha_3 + 4\cdot \alpha_4 &=& 5\cdot \beta \\ 2\cdot \alpha_2 + 3\cdot \alpha_3 + 1\cdot \alpha_1 &=& 5\cdot \beta - 5\cdot \alpha_4 +1\cdot \alpha_4 \\ 1\cdot \alpha_1 - 1\cdot \alpha_4 + 3\cdot \alpha_3 + 2\cdot \alpha_2 &=& 5\cdot (\beta - a_4) \\ 1\cdot \alpha_1 - 1\cdot \alpha_4 + 2\cdot \alpha_2 - 2\cdot \alpha_3 &=& 5\cdot (\beta - a_4 - a_3)\\ \alpha_1 - \alpha_2 + 2(\alpha_2 - \alpha_3) &=&5\gamma \end{matrix}$$
Now I am stuck.
Source: Sbírka příkladů z matematiky pro vysoké školy (1987) - F. Jirásek; the assignment was translated from Czech into English
The key observation is that for every integer $k$ and every choice of $x_0,x_1,x_2,x_3,x_4\in S$ you have $$k\sum x_i+\sum ix_i\equiv \sum ix_{i-k}\pmod{5},$$ if you take the indices mod $5$. So it suffices to find five elements $x_i\in S$ such that $$\sum x_i\not\equiv0\pmod{5},$$ as then this sum is coprime to $5$, and hence you can choose $k$ such that $$\sum ix_{i-k}\equiv k\sum x_i+\sum ix_i\equiv0\pmod{5}.$$ Note that it is not possible to find five such elements if and only if all elements are pairwise congruent mod $5$. Of course in this case any choice of four elements will do, as then $$\sum ix_i\equiv x_1\sum i=x_1(1+2+3+4)\equiv0\pmod{5}.$$