Number the corners of a cube from 1 to 8, so that the sum of the four corners of every face is the same.
If facing the cube from the front, and the back four are in a group, and the front four are a group, then the number are as follows (in their respective order):
back: 67
32
front: 14
85
the faces' numbers would then be: {6,3,1,8}, {7,2,4,5}, {6,7,4,1}, {3,2,5,8}, {6,7,2,3}, and {1,4,5,8}.
This one took me a few minutes. :)
Solution?
