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
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 2002-12-24 23:58:32