An A×B×C rectangular block has been divided into ABC unit cubes. Some of these cubes have faces on the surface of the block, while others are entirely interior.
Show that for any positive integer N there is a block with N times as many interior cubes as surface cubes.
Find the dimensions of such a block having the smallest volume.
So let's start simple and assume the edges are in a sequence; say that the dimensions of the interior block is x, x+2, x+4. Then the full block is x+2, x+4, x+6.
The condition that there are N times as many interior cubes can be rewritten as the ration of interior block to full block is N/(N+1).
Then I can say x/(x+6) = N/(N+1). A bit of algebra then gives us x=6N. Then for any integer N we can say A=6N+2, B=6N+4, and C=6N+6 describes a full block which has N times as many interior cubes as surface cubes.
Solution
