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

1. Aug 24, 2007

### Werg22

Prove or disprove that it is possible.

2. Aug 25, 2007

### Rogerio

Well,the central cube needs 6 cuts.
So, 6 is the minimal number of cuts.

3. Sep 21, 2007

### Xori

Yes, you can use a knife with two blades so you only make one cut in each direction

4. Sep 21, 2007

### Rogerio

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!

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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
5. Sep 22, 2007

### Jimmy Snyder

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.

eom

6. Sep 22, 2007

### AlephZero

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.

7. Sep 24, 2007

### Rogerio

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.

8. Sep 24, 2007

### Xori

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.

9. Oct 14, 2007

### Wild Angel

This seemed interesting so, I do tried doing so and it is very possible.Perfect 27

10. Oct 14, 2007

### akhil982

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. :(

11. Oct 14, 2007

### futurebird

Of course, Am I missing something here?

I mean, how many ways can you make a cube out of 27 smaller cubes anyway?

12. Oct 14, 2007

### akhil982

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

13. Oct 15, 2007

### Wild Angel

Look at a Rubics cube, 4 cuts on top and 2 cuts to the side.

14. Oct 15, 2007

### Sleek

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.

15. Oct 16, 2007

### K.J.Healey

Thats 6 cuts? I think the whole point is less than six cuts.

16. Oct 16, 2007

### Wild Angel

Can you figure out a way with less cuts?

17. Oct 16, 2007

### Rogerio

You just can't - read posts #2 and #7.

18. Oct 16, 2007

### JDEEM

It can be done in 4 cuts. think about it.

19. Oct 16, 2007

### Rogerio

I don't see how.

Could you please explain it to us ?

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Hint: read posts #2 and #7

Last edited: Oct 16, 2007
20. Oct 16, 2007

### Kittel Knight

Parallelepipeds are not necessarily cubes. :yuck: