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Re: A recess in a cube

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  • richard_heylen
    ... I think that the fact that a^3 + b^3 c^3 for integers makes it seem reasonable to me that this assertion is true. I can t however, off the top of my
    Message 1 of 2 , May 7, 2004
      --- In primenumbers@yahoogroups.com, "gulland68" <tmgulland@h...>
      wrote:
      > When a cubic structure is subtracted from a larger cube composed of
      > cubic blocks, the blocks remaining from the larger cube cannot be
      > reassembled into a cube

      I think that the fact that a^3 + b^3 <> c^3 for integers makes it
      seem reasonable to me that this assertion is true. I can't however,
      off the top of my head, rule out the cases in which the ratios of the
      lengths of the cubic blocks are irrational or in which two or more of
      the cubes comprise an infinite number of cubic pieces.

      , since if the subtracted cube is removed in
      > such a way that it an the larger cube share a common vertex, there
      > are invariably two sides of the remains of the host cube that are
      > rectangular if the third is taken to be a square base. Howeverone
      > rearranges the blocks into squares there must be some blocks
      > remaining that can only be arranged as a rectangle or not grouped
      > together as a quadrangle at all.

      I don't fully understand what you mean and what I think I understand,
      I think is wrong. It might be instructive to consider a cube divided
      into 43944934864044921875 equal cubic pieces. You can then remove a
      cube with 1214928 pieces along an edge and rearrange the remainder to
      form a cube with 3480205 pieces along an edge. You then only have 2
      cubic pieces left over!

      Richard
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