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Can a cube be cut in 27 smaller cubes in less than 6 cuts?

  1. Aug 24, 2007 #1
    Prove or disprove that it is possible.
  2. jcsd
  3. Aug 25, 2007 #2
    Well,the central cube needs 6 cuts.
    So, 6 is the minimal number of cuts.

  4. Sep 21, 2007 #3
    Yes, you can use a knife with two blades so you only make one cut in each direction
  5. Sep 21, 2007 #4
    It doesn't matter if you are going to use the knife just 3 times.
    The question is about the number of cuts - and you need 6 cuts!

    And, according to your point of view, why didn't you use acid instead a knife?
    So you would need no cuts at all ! :rofl:
    Last edited: Sep 21, 2007
  6. Sep 22, 2007 #5
    I don't have an answer. However, I see that I can cut it into 64 smaller cubes with 6 cuts, So at least it seems reasonable that you could get 27 in 5 . In any case, we must define 'a cut' in such a way that when I cut two pieces into four with a single action of the knife, that is one cut, not two. Cut as follows:

    First make three cuts one each down the middle of each face and perpendicular to the sides. This will cut the cube into 8 smaller cubes. Take the 8 cubes and lay them out in a row. Cut all 8 cubes down the middle of the row, so that each of the 8 smaller cubes is in the same condition as the larger one was after the very first cut. Then without disturbing the relative positions of the two pieces of any of the smaller split cubes, rearrange the cubes in a row perpendicular to the cut. Cut again down the middle of the row. Now each of the smaller cubes looks like the larger cube did after the second cut. Rearrange and cut a third time. Now you will have 64 cubes.

  7. Sep 22, 2007 #6


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    A different argument - same conclusion as Rogiero:

    The first cut makes two pieces size 1x3x3 and 2x3x3.

    The 2x3x3 piece contains 18 cubes, but the maximum number of pieces you can make in 4 cuts is 2^4 = 16.

  8. Sep 24, 2007 #7
    Unfortunately we can't...

    Well,for the optimal policy,
    it seems reasonable that all the smaller cubes are the same size (in order to take advantage of every cut).
    Then, even employing the optimal policy to get at least 27 cubes, there will be at least one central cube, in such way that none of its faces shares any external face. So, you need one cut to get each face. It means you need exactly 6 cuts to make the central cube. So, 6 is the minimum.

  9. Sep 24, 2007 #8
    I didn't use acid because it won't travel sideways to make a "cut" and I'm too lazy to turn the cube after every "cut" :P

    And I was thinking of a "cut" as the action of the motion of the knife, not the resulting change in shape.
  10. Oct 14, 2007 #9
    This seemed interesting so, I do tried doing so and it is very possible.Perfect 27
  11. Oct 14, 2007 #10
    In simple terms.. if you make a,b and c number of cuts along length,breadth and height , you'll have (a+1)*(b+1)*(c+1)=27. And for a+b+c to be minimum, the optimal solution is a=b=c .. and that leads to a=b=c=2. hence total number of 6 cuts.

    I don't think it can be better than this. But everything is possible. :(
  12. Oct 14, 2007 #11
    Of course, Am I missing something here?

    I mean, how many ways can you make a cube out of 27 smaller cubes anyway?
  13. Oct 14, 2007 #12
    I think you can do that in 3 ways..not counting repeatitions.. cube of 1, cube of 2*2*2 and a cube of 3*3*3
  14. Oct 15, 2007 #13
    Look at a Rubics cube, 4 cuts on top and 2 cuts to the side.
  15. Oct 15, 2007 #14
    Yup, same as Wild Angel. Four cuts on top (with two parallel cuts perpendicular to other two parallel cuts). This makes 9 parallelepiped like things. Two cuts sideways, we have 9 cubes on each of the three layers. Thus, 3*9=27 cubes.
  16. Oct 16, 2007 #15
    Thats 6 cuts? I think the whole point is less than six cuts.
  17. Oct 16, 2007 #16
    Can you figure out a way with less cuts?
  18. Oct 16, 2007 #17
    You just can't - read posts #2 and #7.

  19. Oct 16, 2007 #18
    It can be done in 4 cuts. think about it.
  20. Oct 16, 2007 #19
    I don't see how.

    Could you please explain it to us ?

    Hint: read posts #2 and #7
    Last edited: Oct 16, 2007
  21. Oct 16, 2007 #20
    Parallelepipeds are not necessarily cubes. :yuck:

    Think about it.
  22. Oct 17, 2007 #21
    So, what if you could re-arrange the pieces before making subsequent cuts? Further, just to be clear, I'm assuming a "cut" represents a flat rectangular plane (finite or not), which must start outside the object being "cut".

    In theory, if you allowed re-arrangements, you could create 32 *pieces* with your 5th cut, not all of which will be cubical. Is it possible, however, that 27 or more of them are cubical? I'd guess not, but I'm having a hard time thinking of how to actually prove that fact, bearing in mind that the cubes aren't necessarily the same size.

  23. Oct 17, 2007 #22


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    Actually, what is the maximum number of solid objects you can obtain when you cut an sphere with n planes?
  24. Oct 17, 2007 #23


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    I found two webpages stating related results, I have no idea where to find a systematic collection of cutting problems or theorems.
    http://www.math.toronto.edu/mathnet/SOAR2003/Winter/index.html tells that a torus can be cut in [itex](n^3+3n^2+8n)/6[/itex] pieces for n cuts.

    http://www.cs.colostate.edu/~rmm/mathChallenge/cairoliBinom.pdf tells that a cube can be cut in
    [tex]\begin{pmatrix} n+1 \\ 3 \end{pmatrix} + n +1 [/tex]
    pieces for n cuts.

    and thus with n=5 it is 26 pieces.
    and with n=4 it is 15.
    Last edited: Oct 17, 2007
  25. Oct 17, 2007 #24


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  26. Oct 17, 2007 #25
    Of course you can!
    However the number of cuts remains 6...

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