Difference between revisions of "1997 PMWC Problems/Problem T7"

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The corners of the 5-5-5 must be the corners of the 4-4-4, because they have the most colored faces and there are only 8 of them. The edges of the 4-4-4 are the edges of the 5-5-5 minus one, and the faces of the 4-4-4 are the faces of the 5-5-5 minus 9. Then we can fill in the center of the 4-4-4 with anything and it will work. Thus, <math>\boxed{64}</math> works.
 
The corners of the 5-5-5 must be the corners of the 4-4-4, because they have the most colored faces and there are only 8 of them. The edges of the 4-4-4 are the edges of the 5-5-5 minus one, and the faces of the 4-4-4 are the faces of the 5-5-5 minus 9. Then we can fill in the center of the 4-4-4 with anything and it will work. Thus, <math>\boxed{64}</math> works.
  
==See also==
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==See Also==
 
{{PMWC box|year=1997|num-b=T6|num-a=T8}}
 
{{PMWC box|year=1997|num-b=T6|num-a=T8}}
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[[Category:Introductory Combinatorics Problems]]

Revision as of 16:07, 15 May 2012

Problem

Color the surfaces of a cube of dimension 5*5*5 red, and then cut the cube into smaller cubes of dimension 1*1*1. Take out all the smaller cubes which have at least one red surface and fix a cuboid, keeping the surfaces of the cuboid red. Now what is the maximum possible volume of the cuboid?

Solution

The number of cubes with at least one red face is $5^3-(5-2)^3=125-27=98$. The smallest cube below that is 64, thus we need to prove that we can make a 4-4-4 cube with a totally red surface.

The corners of the 5-5-5 must be the corners of the 4-4-4, because they have the most colored faces and there are only 8 of them. The edges of the 4-4-4 are the edges of the 5-5-5 minus one, and the faces of the 4-4-4 are the faces of the 5-5-5 minus 9. Then we can fill in the center of the 4-4-4 with anything and it will work. Thus, $\boxed{64}$ works.

See Also

1997 PMWC (Problems)
Preceded by
Problem T6
Followed by
Problem T8
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10