Color the cubes, then assemble them to form a larger cube












5














Given 27 same-sized cubes and using three colors of your choice, can you color the faces of the cubes in such a way that it's possible to assemble the cubes into a 3x3x3 cube with only one color on the surface, no matter which of the colors is chosen as the surface color?






geometry






share









New contributor




Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • 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
    3 hours ago












  • You must be able to assemble the cube in each of the colors.
    – Daniel Mathias
    2 hours ago










  • Oh, I see. Thanks for clarifying!
    – Frpzzd
    2 hours ago










  • 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
    44 mins ago
















5














Given 27 same-sized cubes and using three colors of your choice, can you color the faces of the cubes in such a way that it's possible to assemble the cubes into a 3x3x3 cube with only one color on the surface, no matter which of the colors is chosen as the surface color?






geometry






share









New contributor




Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • 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
    3 hours ago












  • You must be able to assemble the cube in each of the colors.
    – Daniel Mathias
    2 hours ago










  • Oh, I see. Thanks for clarifying!
    – Frpzzd
    2 hours ago










  • 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
    44 mins ago














5












5








5







Given 27 same-sized cubes and using three colors of your choice, can you color the faces of the cubes in such a way that it's possible to assemble the cubes into a 3x3x3 cube with only one color on the surface, no matter which of the colors is chosen as the surface color?






geometry






share









New contributor




Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Given 27 same-sized cubes and using three colors of your choice, can you color the faces of the cubes in such a way that it's possible to assemble the cubes into a 3x3x3 cube with only one color on the surface, no matter which of the colors is chosen as the surface color?






geometry




geometry





share









New contributor




Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










share









New contributor




Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








share



share








edited 47 mins ago









Bass

27.1k465168




27.1k465168






New contributor




Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 3 hours ago









Daniel Mathias

261




261




New contributor




Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • 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
    3 hours ago












  • You must be able to assemble the cube in each of the colors.
    – Daniel Mathias
    2 hours ago










  • Oh, I see. Thanks for clarifying!
    – Frpzzd
    2 hours ago










  • 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
    44 mins ago


















  • 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
    3 hours ago












  • You must be able to assemble the cube in each of the colors.
    – Daniel Mathias
    2 hours ago










  • Oh, I see. Thanks for clarifying!
    – Frpzzd
    2 hours ago










  • 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
    44 mins ago
















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
3 hours ago






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
3 hours ago














You must be able to assemble the cube in each of the colors.
– Daniel Mathias
2 hours ago




You must be able to assemble the cube in each of the colors.
– Daniel Mathias
2 hours ago












Oh, I see. Thanks for clarifying!
– Frpzzd
2 hours ago




Oh, I see. Thanks for clarifying!
– Frpzzd
2 hours ago












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
44 mins ago




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
44 mins ago










2 Answers
2






active

oldest

votes


















4














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.






share|improve this answer





















  • Much better than my answer; well done!
    – Hugh
    1 hour ago










  • @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
    1 hour ago










  • 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
    1 hour ago










  • This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
    – JonMark Perry
    39 mins ago



















0














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.







share|improve this answer





















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    2 Answers
    2






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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

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    4














    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.






    share|improve this answer





















    • Much better than my answer; well done!
      – Hugh
      1 hour ago










    • @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
      1 hour ago










    • 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
      1 hour ago










    • This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
      – JonMark Perry
      39 mins ago
















    4














    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.






    share|improve this answer





















    • Much better than my answer; well done!
      – Hugh
      1 hour ago










    • @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
      1 hour ago










    • 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
      1 hour ago










    • This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
      – JonMark Perry
      39 mins ago














    4












    4








    4






    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.






    share|improve this answer












    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.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 2 hours ago









    Frpzzd

    642119




    642119












    • Much better than my answer; well done!
      – Hugh
      1 hour ago










    • @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
      1 hour ago










    • 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
      1 hour ago










    • This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
      – JonMark Perry
      39 mins ago


















    • Much better than my answer; well done!
      – Hugh
      1 hour ago










    • @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
      1 hour ago










    • 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
      1 hour ago










    • This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
      – JonMark Perry
      39 mins ago
















    Much better than my answer; well done!
    – Hugh
    1 hour ago




    Much better than my answer; well done!
    – Hugh
    1 hour ago












    @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
    1 hour ago




    @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
    1 hour ago












    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
    1 hour ago




    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
    1 hour ago












    This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
    – JonMark Perry
    39 mins ago




    This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
    – JonMark Perry
    39 mins ago











    0














    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.







    share|improve this answer


























      0














      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.







      share|improve this answer
























        0












        0








        0






        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.







        share|improve this answer












        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.








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 2 hours ago









        Hugh

        1,3511617




        1,3511617






















            Daniel Mathias is a new contributor. Be nice, and check out our Code of Conduct.










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