Three cubes of volume 1, 8, and 64 are glued together.
What is the smallest possible surface area of the resulting configuration?
Problem taken from UNL Math Day
(In reply to
As easy as 1,2,4 by TomM)
Correction to your math.
First of all, the surfaces areas are 6, 24, 96, for a total of 126.
Secondly, the area would be reduced by 12 (I think you forgot to add 2).
So the minimum area would be 114. Other than these area, I agree completely with your logic (unless there's some obscure way to piece them together).

Posted by np_rt
on 20021224 23:58:32 