## 8 points to the person to solve this puzzle.

(In the spirit of the contest, please refrain from googling answers)

I have a cube with sides of length 3L. I want to use my saw to cut my cube into 27 identical blocks with sides of length of L. The trivial way to do this is with 6 cuts (2 through each face of the cube). If I am allowed to rearrange the pieces after each cut, can I do it in fewer than 6 cuts?

Justify your answer.

Let’s see. I can think of a way to do it with 4 cuts.

2 cuts will give you 3 boxes (I forget the correct term for the geometrical shape that has 6 rectangular faces, but I didn’t google it) of 3Lx3LxL. If you stack those so that 3LxL faces touch, 2 more cuts will give you 27 L-sided cubes.

Sal said this on August 8, 2008 at 4:18 pm |

I take it back; you’d still need 2 more cuts to get 27 cubes. FAIL.

Sal said this on August 8, 2008 at 4:21 pm |

The only way for the inner “center” cube of side L to exist is by cutting 6 clean faces on it. Thus, the minimum number of cuts is 6.

theJenksster said this on August 8, 2008 at 10:22 pm |

Winner winner, chicken dinner.

wcuk said this on August 8, 2008 at 11:36 pm |