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>. | ||
| − | '''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. | + | '''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. |
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.) | 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. | + | 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. Note that <math>8\times4\times3=96</math> satisfies Constraint 1 and <math>8+4+3=15</math> satisfies Constrain 2. Constrain 3 is clearly satisfied as well. So, <math>\boxed{96}</math> works. |
==See Also== | ==See Also== | ||
Revision as of 01:27, 25 January 2020
Problem
Color the surfaces of a cube of dimension 
red, and then cut the cube into smaller cubes of dimension 
. 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 
. Therefore, the volume of the cube cannot more than 
.
Constraint 2: If the dimensions of the cube are 
, 
. This is because the total length of the edges is simply 
, and both have 
corners, so simply divide by 
, to get 
.
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 
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 
satisfies Constraint 1 and 
satisfies Constrain 2. Constrain 3 is clearly satisfied as well. So, 
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 | ||
