Difference between revisions of "1997 PMWC Problems/Problem T7"
(→See also) |
m (→Solution) |
||
(5 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | Color the surfaces of a cube of dimension 5 | + | Color the surfaces of a cube of dimension <math>5\times 5\times 5</math> red, and then cut the cube into smaller cubes of dimension <math>1\times 1\times 1</math>. Take out all the smaller cubes which have at least one red surface and fix a cuboid, <b>keeping the surfaces of the cuboid red</b>. Now what is the maximum possible volume of the cuboid? |
==Solution== | ==Solution== | ||
− | The number of cubes with at least one red face is <math>5^3-(5-2)^3=125-27=98</math>. | + | '''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>. |
− | + | '''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>. | |
− | ==See | + | '''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 <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== | ||
{{PMWC box|year=1997|num-b=T6|num-a=T8}} | {{PMWC box|year=1997|num-b=T6|num-a=T8}} | ||
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] |
Latest revision as of 21:47, 13 July 2023
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 |