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Difference between revisions of "1997 PMWC Problems/Problem T7"
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'''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 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>.
  
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'''Constraint 3:''' The "side" cubes, which are <math>2\left((x-2)(y-2)+(x-2)(z-2)+(y-2)(z-2)\right)</math> in number. This already factors in the edges and corners.
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'''Constraint 3:''' The "side" cubes, which are in number. This already factors in the edges and corners.
  
  

Latest revision as of 20:47, 13 July 2023

Problem

Color the surfaces of a cube of dimension $5\times 5\times 5$ red, and then cut the cube into smaller cubes of dimension $1\times 1\times 1$. 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 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. Note that $8\times4\times3=96$ satisfies Constraint 1 and $8+4+3=15$ satisfies Constrain 2. Constrain 3 is clearly satisfied as well. So, $\boxed{96}$ 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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