Color the cubes, then assemble them to form a larger cube
Goal: Paint 27 cubes using three colors (for example, red, yellow, and blue), so that you can form a 3x3x3 cube with all surfaces in red (for example), a 3x3x3 cube all in yellow, and a 3x3x3 cube all in blue by rearranging the painted cubes, not repainting them.
geometry three-dimensional
New contributor
add a comment |
Goal: Paint 27 cubes using three colors (for example, red, yellow, and blue), so that you can form a 3x3x3 cube with all surfaces in red (for example), a 3x3x3 cube all in yellow, and a 3x3x3 cube all in blue by rearranging the painted cubes, not repainting them.
geometry three-dimensional
New contributor
Welcome to PSE! Your question seems unclear to me. Why not just put the cubes into a 3x3x3 cube shape, paint the faces of the larger cube, disassemble, and then paint the rest of the faces whatever colors we want?
– Frpzzd
Dec 29 '18 at 22:06
1
You must be able to assemble the cube in each of the colors.
– Daniel Mathias
Dec 29 '18 at 22:09
Oh, I see. Thanks for clarifying!
– Frpzzd
Dec 29 '18 at 22:10
1
I ran into the same problem as @Frpzzd, so I tried to edit the question to be a bit more difficult to misunderstand. My sentence structure still seems a bit convoluted, so please feel free to improve it however you see fit.
– Bass
Dec 30 '18 at 0:22
add a comment |
Goal: Paint 27 cubes using three colors (for example, red, yellow, and blue), so that you can form a 3x3x3 cube with all surfaces in red (for example), a 3x3x3 cube all in yellow, and a 3x3x3 cube all in blue by rearranging the painted cubes, not repainting them.
geometry three-dimensional
New contributor
Goal: Paint 27 cubes using three colors (for example, red, yellow, and blue), so that you can form a 3x3x3 cube with all surfaces in red (for example), a 3x3x3 cube all in yellow, and a 3x3x3 cube all in blue by rearranging the painted cubes, not repainting them.
geometry three-dimensional
geometry three-dimensional
New contributor
New contributor
edited Dec 30 '18 at 1:19
Brandon_J
1719
1719
New contributor
asked Dec 29 '18 at 21:57
Daniel Mathias
584
584
New contributor
New contributor
Welcome to PSE! Your question seems unclear to me. Why not just put the cubes into a 3x3x3 cube shape, paint the faces of the larger cube, disassemble, and then paint the rest of the faces whatever colors we want?
– Frpzzd
Dec 29 '18 at 22:06
1
You must be able to assemble the cube in each of the colors.
– Daniel Mathias
Dec 29 '18 at 22:09
Oh, I see. Thanks for clarifying!
– Frpzzd
Dec 29 '18 at 22:10
1
I ran into the same problem as @Frpzzd, so I tried to edit the question to be a bit more difficult to misunderstand. My sentence structure still seems a bit convoluted, so please feel free to improve it however you see fit.
– Bass
Dec 30 '18 at 0:22
add a comment |
Welcome to PSE! Your question seems unclear to me. Why not just put the cubes into a 3x3x3 cube shape, paint the faces of the larger cube, disassemble, and then paint the rest of the faces whatever colors we want?
– Frpzzd
Dec 29 '18 at 22:06
1
You must be able to assemble the cube in each of the colors.
– Daniel Mathias
Dec 29 '18 at 22:09
Oh, I see. Thanks for clarifying!
– Frpzzd
Dec 29 '18 at 22:10
1
I ran into the same problem as @Frpzzd, so I tried to edit the question to be a bit more difficult to misunderstand. My sentence structure still seems a bit convoluted, so please feel free to improve it however you see fit.
– Bass
Dec 30 '18 at 0:22
Welcome to PSE! Your question seems unclear to me. Why not just put the cubes into a 3x3x3 cube shape, paint the faces of the larger cube, disassemble, and then paint the rest of the faces whatever colors we want?
– Frpzzd
Dec 29 '18 at 22:06
Welcome to PSE! Your question seems unclear to me. Why not just put the cubes into a 3x3x3 cube shape, paint the faces of the larger cube, disassemble, and then paint the rest of the faces whatever colors we want?
– Frpzzd
Dec 29 '18 at 22:06
1
1
You must be able to assemble the cube in each of the colors.
– Daniel Mathias
Dec 29 '18 at 22:09
You must be able to assemble the cube in each of the colors.
– Daniel Mathias
Dec 29 '18 at 22:09
Oh, I see. Thanks for clarifying!
– Frpzzd
Dec 29 '18 at 22:10
Oh, I see. Thanks for clarifying!
– Frpzzd
Dec 29 '18 at 22:10
1
1
I ran into the same problem as @Frpzzd, so I tried to edit the question to be a bit more difficult to misunderstand. My sentence structure still seems a bit convoluted, so please feel free to improve it however you see fit.
– Bass
Dec 30 '18 at 0:22
I ran into the same problem as @Frpzzd, so I tried to edit the question to be a bit more difficult to misunderstand. My sentence structure still seems a bit convoluted, so please feel free to improve it however you see fit.
– Bass
Dec 30 '18 at 0:22
add a comment |
4 Answers
4
active
oldest
votes
Paint the small cubes with the colors red, blue, and green as follows, so that on each cube, the faces with the same color are all mutually adjacent.
- One cube with three red faces and three blue faces
- One cube with three red faces and three green faces
- One cube with three blue faces and three green faces
- Three cubes each with three red faces, two blue faces, and one green face
- Three cubes each with three red faces, two green faces, and one blue face
- Three cubes each with three blue faces, two red faces, and one green face
- Three cubes each with three blue faces, two green faces, and one red face
- Three cubes each with three green faces, two red faces, and one blue face
- Three cubes each with three green faces, two blue faces, and one red face
- Six cubes with two faces of each color
Then you have exactly the correct number of corner, edge, face-center, and center pieces for each cube.
Much better than my answer; well done!
– Hugh
Dec 29 '18 at 23:12
@Hugh Thanks! It would be nice to generalize it. The 4x4x4 cube with 3 colors is pretty much trivial, but with 4 colors... hmm...
– Frpzzd
Dec 29 '18 at 23:22
Oh, that's a good puzzle... A 4*4*4 with 3 colours can't be too difficult, but with 4 colours it could get tough. I'm pretty sure it could be done though! You know what I'll be working on ;)
– Hugh
Dec 29 '18 at 23:33
7
This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
– JonMark Perry
Dec 30 '18 at 0:27
1
@Hugh The 4*4*4 with 4 colors is has a similar solution
– Daniel Mathias
Dec 30 '18 at 3:56
|
show 7 more comments
Here is a simple painting procedure that leads to a solution for any size of cube.
- Start with $27$ (or $64$, or $n^3$) unpainted cubes assembled into a large cube.
- Paint the outside faces one colour.
- Rearrange the cubes by removing the bottom layer and placing it on top. Similarly take the left-most slice and place it against the right hand side. Finally take the rear slice and place that against the front face. This puts unpainted surfaces on the outside of the cube.
- Repeat steps 2 and 3 using different colours of paint until all cubes are fully painted.
This method works for any size of cube. Note also that in the finished product, every pair of adjacent cubes has touching faces that share the same colour.
Beautiful stuff
– Jonathan Allan
Dec 30 '18 at 19:46
add a comment |
If we use colours R, G and B, then we can write:
$$(RG+GB+BR)^3$$
which expands into
$$R^3G^3+G^3B^3+B^3R^3+3(R^2G^3B+G^2B^3R+B^2R^3G+R^3G^2B+R^2GB^3+RG^3B^2)+6R^2G^2B^2$$
If we let, w.l.o.g., $R=x, G=B=1$, then we have:
$$(2x+1)^3=8x^3+12x^2+6x+1$$
This describes the number of corner pieces, edge pieces and centre pieces, and the coefficients total $27$.
To explain a motive, we can see that each colour needs $54$ faces, and this fits exactly into the $27times6$ faces we have available.
Therefore the non-visible central cube cannot have any of the colours of the visible colour chose. Therefore there must be three of these, and also as we cannot waste space, they must consist of $2$ sets of $3$ colours. We now need to find six more corners for each colour - these cannot be on the same cube for the same reasons as above, and so we have partially coloured $3times6=18$ more cubes, leaving $6$. These six must have $2$ of each colour, again for the same reasons. The last pieces then fall into place.
Now we can attempt a $4times4times4$ colouring. The 'characteristic polynomial' we are looking for is:
$$8x^3+24x^2+24x+8=(2x+2)^3$$
so we need $4$ (symmetric) elements of two each.
I propose:
$$(RB+GB+BY+YR)^3$$
add a comment |
This sounds very similar to...
this puzzle featured in a video by TED-ED. I'll try and make a picture when I get home.
add a comment |
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4 Answers
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4 Answers
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Paint the small cubes with the colors red, blue, and green as follows, so that on each cube, the faces with the same color are all mutually adjacent.
- One cube with three red faces and three blue faces
- One cube with three red faces and three green faces
- One cube with three blue faces and three green faces
- Three cubes each with three red faces, two blue faces, and one green face
- Three cubes each with three red faces, two green faces, and one blue face
- Three cubes each with three blue faces, two red faces, and one green face
- Three cubes each with three blue faces, two green faces, and one red face
- Three cubes each with three green faces, two red faces, and one blue face
- Three cubes each with three green faces, two blue faces, and one red face
- Six cubes with two faces of each color
Then you have exactly the correct number of corner, edge, face-center, and center pieces for each cube.
Much better than my answer; well done!
– Hugh
Dec 29 '18 at 23:12
@Hugh Thanks! It would be nice to generalize it. The 4x4x4 cube with 3 colors is pretty much trivial, but with 4 colors... hmm...
– Frpzzd
Dec 29 '18 at 23:22
Oh, that's a good puzzle... A 4*4*4 with 3 colours can't be too difficult, but with 4 colours it could get tough. I'm pretty sure it could be done though! You know what I'll be working on ;)
– Hugh
Dec 29 '18 at 23:33
7
This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
– JonMark Perry
Dec 30 '18 at 0:27
1
@Hugh The 4*4*4 with 4 colors is has a similar solution
– Daniel Mathias
Dec 30 '18 at 3:56
|
show 7 more comments
Paint the small cubes with the colors red, blue, and green as follows, so that on each cube, the faces with the same color are all mutually adjacent.
- One cube with three red faces and three blue faces
- One cube with three red faces and three green faces
- One cube with three blue faces and three green faces
- Three cubes each with three red faces, two blue faces, and one green face
- Three cubes each with three red faces, two green faces, and one blue face
- Three cubes each with three blue faces, two red faces, and one green face
- Three cubes each with three blue faces, two green faces, and one red face
- Three cubes each with three green faces, two red faces, and one blue face
- Three cubes each with three green faces, two blue faces, and one red face
- Six cubes with two faces of each color
Then you have exactly the correct number of corner, edge, face-center, and center pieces for each cube.
Much better than my answer; well done!
– Hugh
Dec 29 '18 at 23:12
@Hugh Thanks! It would be nice to generalize it. The 4x4x4 cube with 3 colors is pretty much trivial, but with 4 colors... hmm...
– Frpzzd
Dec 29 '18 at 23:22
Oh, that's a good puzzle... A 4*4*4 with 3 colours can't be too difficult, but with 4 colours it could get tough. I'm pretty sure it could be done though! You know what I'll be working on ;)
– Hugh
Dec 29 '18 at 23:33
7
This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
– JonMark Perry
Dec 30 '18 at 0:27
1
@Hugh The 4*4*4 with 4 colors is has a similar solution
– Daniel Mathias
Dec 30 '18 at 3:56
|
show 7 more comments
Paint the small cubes with the colors red, blue, and green as follows, so that on each cube, the faces with the same color are all mutually adjacent.
- One cube with three red faces and three blue faces
- One cube with three red faces and three green faces
- One cube with three blue faces and three green faces
- Three cubes each with three red faces, two blue faces, and one green face
- Three cubes each with three red faces, two green faces, and one blue face
- Three cubes each with three blue faces, two red faces, and one green face
- Three cubes each with three blue faces, two green faces, and one red face
- Three cubes each with three green faces, two red faces, and one blue face
- Three cubes each with three green faces, two blue faces, and one red face
- Six cubes with two faces of each color
Then you have exactly the correct number of corner, edge, face-center, and center pieces for each cube.
Paint the small cubes with the colors red, blue, and green as follows, so that on each cube, the faces with the same color are all mutually adjacent.
- One cube with three red faces and three blue faces
- One cube with three red faces and three green faces
- One cube with three blue faces and three green faces
- Three cubes each with three red faces, two blue faces, and one green face
- Three cubes each with three red faces, two green faces, and one blue face
- Three cubes each with three blue faces, two red faces, and one green face
- Three cubes each with three blue faces, two green faces, and one red face
- Three cubes each with three green faces, two red faces, and one blue face
- Three cubes each with three green faces, two blue faces, and one red face
- Six cubes with two faces of each color
Then you have exactly the correct number of corner, edge, face-center, and center pieces for each cube.
answered Dec 29 '18 at 22:57
Frpzzd
871120
871120
Much better than my answer; well done!
– Hugh
Dec 29 '18 at 23:12
@Hugh Thanks! It would be nice to generalize it. The 4x4x4 cube with 3 colors is pretty much trivial, but with 4 colors... hmm...
– Frpzzd
Dec 29 '18 at 23:22
Oh, that's a good puzzle... A 4*4*4 with 3 colours can't be too difficult, but with 4 colours it could get tough. I'm pretty sure it could be done though! You know what I'll be working on ;)
– Hugh
Dec 29 '18 at 23:33
7
This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
– JonMark Perry
Dec 30 '18 at 0:27
1
@Hugh The 4*4*4 with 4 colors is has a similar solution
– Daniel Mathias
Dec 30 '18 at 3:56
|
show 7 more comments
Much better than my answer; well done!
– Hugh
Dec 29 '18 at 23:12
@Hugh Thanks! It would be nice to generalize it. The 4x4x4 cube with 3 colors is pretty much trivial, but with 4 colors... hmm...
– Frpzzd
Dec 29 '18 at 23:22
Oh, that's a good puzzle... A 4*4*4 with 3 colours can't be too difficult, but with 4 colours it could get tough. I'm pretty sure it could be done though! You know what I'll be working on ;)
– Hugh
Dec 29 '18 at 23:33
7
This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
– JonMark Perry
Dec 30 '18 at 0:27
1
@Hugh The 4*4*4 with 4 colors is has a similar solution
– Daniel Mathias
Dec 30 '18 at 3:56
Much better than my answer; well done!
– Hugh
Dec 29 '18 at 23:12
Much better than my answer; well done!
– Hugh
Dec 29 '18 at 23:12
@Hugh Thanks! It would be nice to generalize it. The 4x4x4 cube with 3 colors is pretty much trivial, but with 4 colors... hmm...
– Frpzzd
Dec 29 '18 at 23:22
@Hugh Thanks! It would be nice to generalize it. The 4x4x4 cube with 3 colors is pretty much trivial, but with 4 colors... hmm...
– Frpzzd
Dec 29 '18 at 23:22
Oh, that's a good puzzle... A 4*4*4 with 3 colours can't be too difficult, but with 4 colours it could get tough. I'm pretty sure it could be done though! You know what I'll be working on ;)
– Hugh
Dec 29 '18 at 23:33
Oh, that's a good puzzle... A 4*4*4 with 3 colours can't be too difficult, but with 4 colours it could get tough. I'm pretty sure it could be done though! You know what I'll be working on ;)
– Hugh
Dec 29 '18 at 23:33
7
7
This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
– JonMark Perry
Dec 30 '18 at 0:27
This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
– JonMark Perry
Dec 30 '18 at 0:27
1
1
@Hugh The 4*4*4 with 4 colors is has a similar solution
– Daniel Mathias
Dec 30 '18 at 3:56
@Hugh The 4*4*4 with 4 colors is has a similar solution
– Daniel Mathias
Dec 30 '18 at 3:56
|
show 7 more comments
Here is a simple painting procedure that leads to a solution for any size of cube.
- Start with $27$ (or $64$, or $n^3$) unpainted cubes assembled into a large cube.
- Paint the outside faces one colour.
- Rearrange the cubes by removing the bottom layer and placing it on top. Similarly take the left-most slice and place it against the right hand side. Finally take the rear slice and place that against the front face. This puts unpainted surfaces on the outside of the cube.
- Repeat steps 2 and 3 using different colours of paint until all cubes are fully painted.
This method works for any size of cube. Note also that in the finished product, every pair of adjacent cubes has touching faces that share the same colour.
Beautiful stuff
– Jonathan Allan
Dec 30 '18 at 19:46
add a comment |
Here is a simple painting procedure that leads to a solution for any size of cube.
- Start with $27$ (or $64$, or $n^3$) unpainted cubes assembled into a large cube.
- Paint the outside faces one colour.
- Rearrange the cubes by removing the bottom layer and placing it on top. Similarly take the left-most slice and place it against the right hand side. Finally take the rear slice and place that against the front face. This puts unpainted surfaces on the outside of the cube.
- Repeat steps 2 and 3 using different colours of paint until all cubes are fully painted.
This method works for any size of cube. Note also that in the finished product, every pair of adjacent cubes has touching faces that share the same colour.
Beautiful stuff
– Jonathan Allan
Dec 30 '18 at 19:46
add a comment |
Here is a simple painting procedure that leads to a solution for any size of cube.
- Start with $27$ (or $64$, or $n^3$) unpainted cubes assembled into a large cube.
- Paint the outside faces one colour.
- Rearrange the cubes by removing the bottom layer and placing it on top. Similarly take the left-most slice and place it against the right hand side. Finally take the rear slice and place that against the front face. This puts unpainted surfaces on the outside of the cube.
- Repeat steps 2 and 3 using different colours of paint until all cubes are fully painted.
This method works for any size of cube. Note also that in the finished product, every pair of adjacent cubes has touching faces that share the same colour.
Here is a simple painting procedure that leads to a solution for any size of cube.
- Start with $27$ (or $64$, or $n^3$) unpainted cubes assembled into a large cube.
- Paint the outside faces one colour.
- Rearrange the cubes by removing the bottom layer and placing it on top. Similarly take the left-most slice and place it against the right hand side. Finally take the rear slice and place that against the front face. This puts unpainted surfaces on the outside of the cube.
- Repeat steps 2 and 3 using different colours of paint until all cubes are fully painted.
This method works for any size of cube. Note also that in the finished product, every pair of adjacent cubes has touching faces that share the same colour.
answered Dec 30 '18 at 4:40
Jaap Scherphuis
14.8k12565
14.8k12565
Beautiful stuff
– Jonathan Allan
Dec 30 '18 at 19:46
add a comment |
Beautiful stuff
– Jonathan Allan
Dec 30 '18 at 19:46
Beautiful stuff
– Jonathan Allan
Dec 30 '18 at 19:46
Beautiful stuff
– Jonathan Allan
Dec 30 '18 at 19:46
add a comment |
If we use colours R, G and B, then we can write:
$$(RG+GB+BR)^3$$
which expands into
$$R^3G^3+G^3B^3+B^3R^3+3(R^2G^3B+G^2B^3R+B^2R^3G+R^3G^2B+R^2GB^3+RG^3B^2)+6R^2G^2B^2$$
If we let, w.l.o.g., $R=x, G=B=1$, then we have:
$$(2x+1)^3=8x^3+12x^2+6x+1$$
This describes the number of corner pieces, edge pieces and centre pieces, and the coefficients total $27$.
To explain a motive, we can see that each colour needs $54$ faces, and this fits exactly into the $27times6$ faces we have available.
Therefore the non-visible central cube cannot have any of the colours of the visible colour chose. Therefore there must be three of these, and also as we cannot waste space, they must consist of $2$ sets of $3$ colours. We now need to find six more corners for each colour - these cannot be on the same cube for the same reasons as above, and so we have partially coloured $3times6=18$ more cubes, leaving $6$. These six must have $2$ of each colour, again for the same reasons. The last pieces then fall into place.
Now we can attempt a $4times4times4$ colouring. The 'characteristic polynomial' we are looking for is:
$$8x^3+24x^2+24x+8=(2x+2)^3$$
so we need $4$ (symmetric) elements of two each.
I propose:
$$(RB+GB+BY+YR)^3$$
add a comment |
If we use colours R, G and B, then we can write:
$$(RG+GB+BR)^3$$
which expands into
$$R^3G^3+G^3B^3+B^3R^3+3(R^2G^3B+G^2B^3R+B^2R^3G+R^3G^2B+R^2GB^3+RG^3B^2)+6R^2G^2B^2$$
If we let, w.l.o.g., $R=x, G=B=1$, then we have:
$$(2x+1)^3=8x^3+12x^2+6x+1$$
This describes the number of corner pieces, edge pieces and centre pieces, and the coefficients total $27$.
To explain a motive, we can see that each colour needs $54$ faces, and this fits exactly into the $27times6$ faces we have available.
Therefore the non-visible central cube cannot have any of the colours of the visible colour chose. Therefore there must be three of these, and also as we cannot waste space, they must consist of $2$ sets of $3$ colours. We now need to find six more corners for each colour - these cannot be on the same cube for the same reasons as above, and so we have partially coloured $3times6=18$ more cubes, leaving $6$. These six must have $2$ of each colour, again for the same reasons. The last pieces then fall into place.
Now we can attempt a $4times4times4$ colouring. The 'characteristic polynomial' we are looking for is:
$$8x^3+24x^2+24x+8=(2x+2)^3$$
so we need $4$ (symmetric) elements of two each.
I propose:
$$(RB+GB+BY+YR)^3$$
add a comment |
If we use colours R, G and B, then we can write:
$$(RG+GB+BR)^3$$
which expands into
$$R^3G^3+G^3B^3+B^3R^3+3(R^2G^3B+G^2B^3R+B^2R^3G+R^3G^2B+R^2GB^3+RG^3B^2)+6R^2G^2B^2$$
If we let, w.l.o.g., $R=x, G=B=1$, then we have:
$$(2x+1)^3=8x^3+12x^2+6x+1$$
This describes the number of corner pieces, edge pieces and centre pieces, and the coefficients total $27$.
To explain a motive, we can see that each colour needs $54$ faces, and this fits exactly into the $27times6$ faces we have available.
Therefore the non-visible central cube cannot have any of the colours of the visible colour chose. Therefore there must be three of these, and also as we cannot waste space, they must consist of $2$ sets of $3$ colours. We now need to find six more corners for each colour - these cannot be on the same cube for the same reasons as above, and so we have partially coloured $3times6=18$ more cubes, leaving $6$. These six must have $2$ of each colour, again for the same reasons. The last pieces then fall into place.
Now we can attempt a $4times4times4$ colouring. The 'characteristic polynomial' we are looking for is:
$$8x^3+24x^2+24x+8=(2x+2)^3$$
so we need $4$ (symmetric) elements of two each.
I propose:
$$(RB+GB+BY+YR)^3$$
If we use colours R, G and B, then we can write:
$$(RG+GB+BR)^3$$
which expands into
$$R^3G^3+G^3B^3+B^3R^3+3(R^2G^3B+G^2B^3R+B^2R^3G+R^3G^2B+R^2GB^3+RG^3B^2)+6R^2G^2B^2$$
If we let, w.l.o.g., $R=x, G=B=1$, then we have:
$$(2x+1)^3=8x^3+12x^2+6x+1$$
This describes the number of corner pieces, edge pieces and centre pieces, and the coefficients total $27$.
To explain a motive, we can see that each colour needs $54$ faces, and this fits exactly into the $27times6$ faces we have available.
Therefore the non-visible central cube cannot have any of the colours of the visible colour chose. Therefore there must be three of these, and also as we cannot waste space, they must consist of $2$ sets of $3$ colours. We now need to find six more corners for each colour - these cannot be on the same cube for the same reasons as above, and so we have partially coloured $3times6=18$ more cubes, leaving $6$. These six must have $2$ of each colour, again for the same reasons. The last pieces then fall into place.
Now we can attempt a $4times4times4$ colouring. The 'characteristic polynomial' we are looking for is:
$$8x^3+24x^2+24x+8=(2x+2)^3$$
so we need $4$ (symmetric) elements of two each.
I propose:
$$(RB+GB+BY+YR)^3$$
answered Dec 30 '18 at 11:33
JonMark Perry
17.5k63584
17.5k63584
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This sounds very similar to...
this puzzle featured in a video by TED-ED. I'll try and make a picture when I get home.
add a comment |
This sounds very similar to...
this puzzle featured in a video by TED-ED. I'll try and make a picture when I get home.
add a comment |
This sounds very similar to...
this puzzle featured in a video by TED-ED. I'll try and make a picture when I get home.
This sounds very similar to...
this puzzle featured in a video by TED-ED. I'll try and make a picture when I get home.
answered Dec 29 '18 at 22:42
Hugh
1,4581617
1,4581617
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add a comment |
Daniel Mathias is a new contributor. Be nice, and check out our Code of Conduct.
Daniel Mathias is a new contributor. Be nice, and check out our Code of Conduct.
Daniel Mathias is a new contributor. Be nice, and check out our Code of Conduct.
Daniel Mathias is a new contributor. Be nice, and check out our Code of Conduct.
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Welcome to PSE! Your question seems unclear to me. Why not just put the cubes into a 3x3x3 cube shape, paint the faces of the larger cube, disassemble, and then paint the rest of the faces whatever colors we want?
– Frpzzd
Dec 29 '18 at 22:06
1
You must be able to assemble the cube in each of the colors.
– Daniel Mathias
Dec 29 '18 at 22:09
Oh, I see. Thanks for clarifying!
– Frpzzd
Dec 29 '18 at 22:10
1
I ran into the same problem as @Frpzzd, so I tried to edit the question to be a bit more difficult to misunderstand. My sentence structure still seems a bit convoluted, so please feel free to improve it however you see fit.
– Bass
Dec 30 '18 at 0:22