Number the corners of a cube from 1 to 8, so that the sum of the four corners of every face is the same.
The solution has been posted but perhaps some people won't know how to get there. So, here is an easy way to work out this puzzle:
When assigning numbers to corners you should try to balance diametrically opposite sides  eg put 8 on one corner (front face)and on the diametrically opposite corner a 7(rear face), then put the 1 on a corner on the front face, to balance this put a 2 on a corner on the rear face. Carry on similarly with the rest of the numbers.
You'll end up with something like this:
Front face starting at top left hand corner working clockwise: 8 1 5 4
Rear face starting at top left hand corner working clockwise: 6 3 7 2
The faces don't add up so try to balance in this case all faces =18 except two of them, which come to 20 and 16. ie the 20 face needs to lose 2 points and the 16 face has to gain 2 points.
This is easily achieved by swapping the 6 with the 3 and swapping the 4 with the 5.
Now all faces have a sum of 18. QED

Posted by MSH
on 20060324 14:43:48 