\section*{Task 2 - 151022}

Hubert has three boxes, each containing a number of balls.

He takes from the first box and puts into each of the other two as many balls as are already in each of them. Then he takes from the second box and puts into each of the other two as many balls as are currently in each of them. Finally, he takes from the third box and puts into each of the other two as many balls as are currently in each of them.

Afterwards, he finds that each of the boxes contains exactly 64 balls.

Determine the number of balls that each of the boxes originally contained!
}

At the beginning, there are \(a\) balls in the first box, \(b\) balls in the second, and \(c\) balls in the third. The three operations change the contents of the boxes as follows:

\begin{tabular}{lccc} 
Action & 1st Box & 2nd Box & 3rd Box \\
& \(a\) & \(b\) & \(c\) \\
I & \(a-b-c\) & \(2 b\) & \(2 c\) \\
II & \(2 a-2 b-2 c\) & \(3 b-a-c\) & \(4 c\) \\
III & \(4 a-4 b-4 c\) & \(6 b-2 a-2 c\) & \(-a-b+7 c\)
\end{tabular}

All boxes then contain 64 balls. This results in the system of equations

\[
4 a-4 b-4 c=64 \quad-2 a+6 b-2 c=64 \quad-a-b+7 c=64
\]

with the solution \(a=104, b=56\) and \(c=32\). Initially, there were 104 balls in the first box, 56 in the second, and 32 in the third.