Author |
Topic: 218 Cubelets (Read 480 times) |
|
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
The dewdrop slides into the shining Sea
Gender: 
Posts: 4489
|
 |
218 Cubelets
« on: Aug 7th, 2010, 4:05pm » |
 Quote  Modify
|
n3 unit cubelets are assembled to create a larger cube. Some (but not all) of the faces of the cube are then painted. When the cube is dismantled it is discovered that 218 cubelets have paint on them. What is the value of n?
|
|
 IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
MathsForFun
Full Member
Posts: 153
|
|
Re: 218 Cubelets
« Reply #1 on: Aug 7th, 2010, 4:26pm » |
Quote Modify
|
At first glance, this appears to be impossible. For all x in {1..6}, 218/x does not result in a square integer.
|
| « Last Edit: Aug 8th, 2010, 12:35am by MathsForFun » |
IP Logged |
Everything is interesting if you look at it closely enough
|
|
|
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489
|
|
Re: 218 Cubelets
« Reply #2 on: Aug 7th, 2010, 4:35pm » |
Quote Modify
|
on Aug 7th, 2010, 4:26pm, MathsForFun wrote:| At first glance, this appears to be impossible. |
|
Usually the sign of a good puzzle. 
|
| « Last Edit: Aug 8th, 2010, 4:22pm by ThudnBlunder » |
IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: 218 Cubelets
« Reply #3 on: Aug 8th, 2010, 8:05am » |
Quote Modify
|
n=8 and two neighbouring faces of the large cube went unpainted
I just tried all ways the faces on the large cube might be (not all) painted.
For a single face painted: n2 cubelets have paint
For two opposite faces painted: 2n2
For two neighbouring faces: n2+n(n-1)
For three neighbouring faces not coming together at a point painted: n2+2n(n-1)
For three neighbouring faces coming together at a point painted: n2+n(n-1)+(n-1)2
For four faces going round the cube painted: n2+2n(n-1)+n*(n-2)
For two neighbouring faces paint-free: n2+2n(n-1)+(n-1)*(n-2)
For one face paint-free: n2+2n(n-1)+n*(n-2)+(n-2)2
|
| « Last Edit: Aug 8th, 2010, 8:06am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
MathsForFun
Full Member
Posts: 153
|
|
Re: 218 Cubelets
« Reply #4 on: Aug 9th, 2010, 1:24am » |
Quote Modify
|
My program gives the same result as TOWR:
Code:for size: 3 thru 14 do
for sides: 2 thru 5 do (
if (sides = 2) then (
edges: [1],
corners: [[2, 0]] /* [2 sides painted, 3 sides painted] */
)
else if (sides = 3) then (
edges: [2, 3],
corners: [[4, 0], [3, 1]]
)
else if (sides = 4) then (
edges: [4, 5],
corners:[[8, 0], [4, 2]]
)
else if (sides = 5) then (
edges: [8],
corners:[[4, 4]]
),
for edge in edges do
for corner in corners do (
extras: (edge * (size - 2)) + corner[1] + (2 * corner[2]),
if integerp(sqrt(218 + extras) / sides) then
if (size^2 * sides - extras = 218) then
print("Size:", size, " Sides:", sides, " Edge:", edge, " Corner:", corner)
)
)$ |
|
The output is as follows:
Size: 8 Sides: 4 Edge: 5 Corner: [8,0]
Size: 8 Sides: 4 Edge: 5 Corner: [4,2]
Of these, the first is not possible, because if there were 5 shared edges (edges with paint on 2 sides of the cubelets) then there would be fewer than 8 corners with only 2 sides painted.
This appears, then, to confirm Towr's solution.
|
|
IP Logged |
Everything is interesting if you look at it closely enough
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: 218 Cubelets
« Reply #5 on: Aug 9th, 2010, 1:58am » |
Quote Modify
|
A simple way to count the painted cubes is to count the unpainted cubes.
For a size n
0 face painted: n*n*n
1 face painted: (n-1)*n*n
2 faces painted: (n-2)*n*n or (n-1)*(n-1)*n
3 faces painted: (n-2)*(n-1)*n or (n-1)*(n-1)*(n-1)
4 faces painted: (n-2)*(n-2)*n or (n-2)*(n-1)*(n-1)
5 faces painted: (n-2)*(n-2)*(n-1)
6 faces painted: (n-2)*(n-2)*(n-2)
The painted cubes are (n*n*n) minus the above.
|
| « Last Edit: Aug 9th, 2010, 1:59am by Grimbal » |
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print