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

Revision as of 05:03, 10 October 2013

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

Constraint 1: The number of cubes with at least one red face is $5^3-(5-2)^3=125-27=98$ . Therefore, the volume of the cube cannot more than $98$ .

Constraint 2: If the dimensions of the cube are $x-y-z$ , $x+y+z\le15$ . This is because the total length of the edges is simply $4(x+y+z)$ , and both have $8$ corners, so simply divide by $4$ , to get $5+5+5=15$ .

Constraint 3: The "side" cubes, which are $(x-2)(y-2)+(x-2)(z-2)+(y-2)(z-2)$ in number. This already factors in the edges and corners.

The corners of the cuboid are not a constraint. (If you think a cuboid can have more than 8 corners, then you have no business messing with this page. If you do, you are either very bad at mathematics, good enough to contemplate alternate geometries in 3D or crazy.)

A $5\times5\times5$ cube would obviously be impossible. 124 would be the next composite number, but it's highest prime factor is 31. Anyway, it's more than 98. 98 itself wouldn't do, similarly due to it's edges. 97 also wouldn't do. 96 for its sides. And so on... ...until 80. The $5\times4\times4$ cuboid. So, $\boxed{80}$ works.

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 ==Solution==
-The number of cubes with at least one red face is <math>5^3-(5-2)^3=125-27=98</math>. 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.
+'''Constraint 1:''' The number of cubes with at least one red face is <math>5^3-(5-2)^3=125-27=98</math>. Therefore, the volume of the cube cannot more than <math>98</math>.
-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.
+'''Constraint 2:''' If the dimensions of the cube are <math>x-y-z</math>, <math>x+y+z\le15</math>. This is because the total length of the edges is simply <math>4(x+y+z)</math>, and both have <math>8</math> corners, so simply divide by <math>4</math>, to get <math>5+5+5=15</math>.
+'''Constraint 3:''' The "side" cubes, which are <math>(x-2)(y-2)+(x-2)(z-2)+(y-2)(z-2)</math> in number. This already factors in the edges and corners.
+The corners of the cuboid are not a constraint. (If you think a cuboid can have more than 8 corners, then you have no business messing with this page. If you do, you are either very bad at mathematics, good enough to contemplate alternate geometries in 3D or crazy.)
+A <math>5\times5\times5</math> cube would obviously be impossible. 124 would be the next composite number, but it's highest prime factor is 31. Anyway, it's more than 98. 98 itself wouldn't do, similarly due to it's edges. 97 also wouldn't do. 96 for its sides. And so on...     ...until 80. The <math>5\times4\times4</math> cuboid. So, <math>\boxed{80}</math> 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

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

Revision as of 05:03, 10 October 2013

Problem

Solution

See Also