First solution: The result is $3n^2+3n +1 = (n + 1)^3- n^3$ is striking and allows a
geometrical interpretation. One person gets x+y+z = 3n objects with 0 ≤x, y, z≤2n. These are triangular coordinates for an equilateral triangle with altitude 3n.x, y, z
can be interpreted as lattice points (make a figure). The hexagon in the figure can be
interpreted as the projection of the cube with edge n + 1 from which a cube of edge
n is subtracted. This solution is due to Martin Härterich, a gold medallist of the IMO