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 2002-12-24 23:58:32