## Task 1 - 250821

Determine the twelve consecutive integers that have the property that the sum of the two largest of these numbers is equal to the sum of the remaining ten!
Let $x$ be the smallest of the twelve numbers. Then, according to the problem statement,

$$
\begin{gathered}
x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9=x+10+x+11 \\
10 x+45=2 x+21 \quad \Rightarrow \quad x=-3
\end{gathered}
$$

Indeed, the sum of the ten consecutive integers from -3 to 6 is equal to the sum of the two following numbers 7 and 8.

The conditions of the problem are thus exactly satisfied by the numbers -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, and 8.