is following step right proving?

it is first time I prove something so I am wonder whether it is right or wrong.

---given problem---

there is a cube whose one side is 3 inch long.

a person cuts 6 times for making 27 cubes whose one side is 1 inch long.

is there a way to cut less than 6 times for making 27 cubes?

if not, prove it.


for any shape of object,

for making one or more cube, I should cut 6 times

because only one plane would be made for each cut.

and cube have 6 plane.

making 27 cubes is more than making a cube.

so making 27 cubes require at least cutting 6 times which time is times of cutting for making a cube.

I tried to prove "for making one or more cube, it require cutting at least 6 times"

does it same as proving "there is no way to make 27 cube with cutting less than 6 times"?


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    $\begingroup$ If you are referring to the little cube inside, then this sounds good to me. That cube doesn't share any of its faces with the big 3x3x3 cube, and a single cut cannot produce more than one face, as you said (I think?) $\endgroup$ – Jyrki Lahtonen Jul 22 '11 at 8:58
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    $\begingroup$ If I want to make 8 cubes, I only need to cut 3 times. But your argument would still suggest cutting 6 times because "making 8 cubes is more than making a cube". So there is something missing in your argument. $\endgroup$ – Rahul Jul 22 '11 at 9:01
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    $\begingroup$ (And the missing thing is in Jyrki's comment, which I didn't see before posting...) $\endgroup$ – Rahul Jul 22 '11 at 9:02
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    $\begingroup$ The way I saw Jyrki's comment expressed originally was to think of painting the exterior of the original cube. The inner cube has no paint, so needs 6 cuts, no matter how you rearrange the pieces between cuts. This distinguishes the 3x3x3 case from the 2x2x2 case. $\endgroup$ – Ross Millikan Jul 22 '11 at 15:30
  • $\begingroup$ I quite like Ross's use of paint here. Admittedly my choice of phrase "shares a face" is not quite as precise as we might like it to be, but at least you understood. $\endgroup$ – Jyrki Lahtonen Jul 23 '11 at 7:43

The question can be rephrased like this: If I have an initial cube, can I cut it in 27 cubes with less than $6$ cuts?

Let's see how many "lego tiles" you can do with $6$ cuts, and we will proceed from this point forward (less cuts etc...).

$a$ cuts on one face produce $a+1$ parts. If you cut each face $a,b,c$ times respectively, with $a+b+c=6$; you get $(a+1)(b+1)(c+1)$ lego tiles.

Since $6$ is not too high a number, it is easy to do all the cases by hand, given also that the solution is identical by rotation of $a,b,c$ (which in itself is a hint).

The maximum is given by $a=b=c=2$, which gives $27$ tiles.

If you have one less cut, from the initial function $(a+1)(b+1)(c+1)$, you see that you can't get $27$ parts...

Of course you could do a more complex demonstration with the use of the gradient of a multiple variable function with linked extrema, but I think this is overkill here.

  • $\begingroup$ Are you sure? I can make 64 cubes with six cuts. Cut twice in each direction - the second time cutting through (appropriately piled on top of each other) both halfs created by the first cut. $\endgroup$ – Jyrki Lahtonen Nov 5 '14 at 17:39
  • $\begingroup$ You are adding in that case a new step, which is not described, and probably not considered, which is piling up parts. $\endgroup$ – Martigan Nov 5 '14 at 17:51

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