# what is the least possible number of cuts required to cut a cube into 27 identical pieces?

### Mohammed

Guys, does anyone know the answer?

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**snags697**

Newbie Mad Physicist Gender: Posts: 14

**27 cubes (big hint)**

« ** on:** Aug 1st, 2002, 12:42pm » Quote Modify

To figure out how many cuts are required to divide a single cube into 27 equal small cubes, think about the center small cube.

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**mook**

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**Re: 27 cubes (big hint)**

« **Reply #1 on:** Aug 3rd, 2002, 8:09am » Quote Modify Remove

just get out paper and scissors and see how many cuts it takes

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**william wu**

**wu::riddles Administrator**

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**Re: 27 cubes (big hint)**

« **Reply #2 on:** Aug 3rd, 2002, 6:39pm » Quote Modify

**on Aug 3rd, 2002, 8:09am, mook wrote:**

just get out paper and scissors and see how many cuts it takes

well the question is asking what the *minimum* number of cuts is, so playing with scisssors may not really help. snags's hint is a good one.

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**ML**

Newbie Gender: Posts: 1

**Re: 27 cubes (big hint)**

« **Reply #3 on:** Aug 7th, 2002, 3:52pm » Quote Modify

Is it possible to do this with less than 5 cuts?

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**Jeremiah Smith**

Full Member Beep! Posts: 172

**Re: 27 cubes (big hint)**

« **Reply #4 on:** Aug 8th, 2002, 5:04pm » Quote Modify

I'm a little curious as to how you would do it with *exactly* 5, much less *less* than 5.

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**Kip Lubliner**

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**Re: 27 cubes (big hint)**

« **Reply #5 on:** Aug 13th, 2002, 11:17am » Quote Modify Remove

I came up with a solution different than snag's. For a hint about my solution, here is a proof that we need at least 5 cuts:

We start with 1 object. Each cut can, at best, double the number of objects. So after 1 cut we are left with, at most, 2 objects. After 2 cuts we are left with, at most, 4 objects. Continuing in this manner, after 4 cuts we have, at most, 2^4 = 16 objects. 16 < 27, so we need more than 4 cuts.

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**snags697**

Newbie Mad Physicist Gender: Posts: 14

**Re: 27 cubes (big hint)**

« **Reply #6 on:** Aug 13th, 2002, 12:07pm » Quote Modify

The solution:

You can do it in a minimum of 6 cuts. Each face of the center cube must be cut once, and you can't possibly cut 2 faces of the same cube at the same time. Therefore, you need all 6 cuts to separate the 27 cubes.

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**NickH**

Senior Riddler Gender: Posts: 341

**Re: 27 cubes**

« **Reply #7 on:** Aug 25th, 2002, 4:49pm » Quote Modify

"A cube is to be cut into 27 smaller cubes (just like a Rubik's Cube). It is clear that this can be done with 6 cuts to the original cube (2 in the x, 2 in the y, 2 in the z). Now, assuming that you can arrange the pieces however you like before doing a cut, what is the minimum number of cuts required to obtain the 27 smaller cubes? Prove your answer."

Three cuts are necessary.

The first cut is constrained, since the block is in one piece. Before the second cut we topple the block of 9 and place it under the 18. We cut through the block of 18 and then begin to cut the block of 9. While doing this, we topple the newly cut two blocks of 9 and place them under the first block of 9. I won't try to describe the rest of the actions, but I think the point is clear that we can continue to remove partly cut blocks, rearrange them as necessary, and feed them in the other end. This is legitimate because it takes place "before doing a cut," in fact, before the third and final cut.

The second cut, then, which is continuous, does most of the work. The third cut would not actually be needed if the question stated that you could rearrange the pieces _after_ a cut! In that case, we could do all the necessary rearranging "after the first cut."

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

Prove or disprove that it is possible.

Forums The Lounge General Discussion

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

Thread starter Werg22 Start date Aug 24, 2007 Aug 24, 2007 #1 Werg22

Prove or disprove that it is possible.

## Answers and Replies

Aug 25, 2007 #2 Rogerio Werg22 said:

Prove or disprove that it is possible.

Well,the central cube needs 6 cuts.

So, 6 is the minimal number of cuts.

Sep 21, 2007 #3 Xori

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

Sep 21, 2007 #4 Rogerio Xori said:

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

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

Sep 22, 2007 #5 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 Sep 22, 2007 #6 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.

Sep 24, 2007 #7 Rogerio jimmysnyder said:

... it seems reasonable that you could get 27 in 5 .

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.

Sep 24, 2007 #8 Xori Rogerio said:

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:

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.

Oct 14, 2007 #9

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

Oct 14, 2007 #10 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. :(

Oct 14, 2007 #11 futurebird

Of course, Am I missing something here?

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

Oct 14, 2007 #12 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

Oct 15, 2007 #13

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

Oct 15, 2007 #14 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.

Oct 16, 2007 #15 K.J.Healey

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

Oct 16, 2007 #16

Can you figure out a way with less cuts?

Oct 16, 2007 #17 Rogerio Wild Angel said:

Can you figure out a way with less cuts?

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

Oct 16, 2007 #18 JDEEM

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

Oct 16, 2007 #19 Rogerio JDEEM said:

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

I don't see how.

Could you please explain it to us ?

-------

Hint: read posts #2 and #7

Last edited: Oct 16, 2007

## In how many minimum cuts can you cut a cube into 27 pieces?

Answer (1 of 5): I don’t know the answer, but the many answers that you’ve already gotten answer it incorrectly, as they assume that the cuts are made orthogonally. If we simplify the problem to that of a square being cut into pieces, it is easy to see that we can get more pieces with fewer cuts ...

In how many minimum cuts can you cut a cube into 27 pieces?

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Sort Siddhartha Chhabra

computer science engineer.6y

In 6 cuts,

2 cuts on each any three sides sharing a corner.

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Anirban Acharya

Eat, sleep, code, repeat!!6y

6 is the answer. The minimum number of cuts can be achieved when you cut the cube in equally length-wise, breadth-wise, height-wise. If you can adjust the position of each new cuts keeping them parallel to each other in their respective plane, you’ll see 3 cuts give 8 pieces, 6 gives 27, 9 gives 64 and so on. It is happening as (

3√n−1)x3 3√n−1)x3

We subtract 1 since for x cut you get x+1 pieces in every plane. We got 3 planes hence x3. Thus we get a relation as

f(n)=(3√n−1)x3 f(n)=(3√n−1)x3

for n being proper cubes. Well that shouldn’t be too hard to code up. The follow-up question is In how many minimum cuts can y

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Aaron Jantzen

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What is the maximum number of identical pieces a cube can be cut into by 4 cuts?

Any Japanese chef can give you a far better answer to this than your typical mathematician will. A mathematician might tell you 12 or 16 depending upon whether they rearrange the pieces before cutting. However a Japanese chef will tell you “thousands” and that is meant literally. Not only can it be done, it is indeed regularly done by these chefs. The secret is called the Katsuramuki technique. Quite simply, the first cut is a spiral, which is normally done by hand with a sharp knife on a log-shaped vegetable such as Daikon, resulting in a long sheet, but could work with a cube by gradually ro

James Barton

Physics Amateur, Read 50+ Physics Books, Understood MostAuthor has 3.2K answers and 11.6M answer views2y

I don’t know the answer, but the many answers that you’ve already gotten answer it incorrectly, as they assume that the cuts are made orthogonally. If we simplify the problem to that of a square being cut into pieces, it is easy to see that we can get more pieces with fewer cuts by increasing the number of intersections - simply by making the cuts at angles that cross a higher number of other cuts.

If this were extended to three dimensions (a cube) then the number of potential intersections increases greatly. I don’t know if it can be done in less than six cuts, but the existing answers propose

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Anonymous 9y

let the cube dimension be 'A'

make a cut at A/3 and 2A/3 on length, breadth and height.

count=0;

for(i=A/3;i<=2*A/3;i+=A/3)

count++;

//now along one dimension is complete...

count*=3;// now count value is the answer

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Manohar Reddy Poreddy

Ramanujan in 60 days; 0 to Top100 Algo in 130d; Top400 DS 40Author has 677 answers and 3.5M answer viewsUpdated 6y

Most obvious looking one is:

1. From the top side, cut horizontally 2 times

2. Again, From the top side, cut vertically 2 times

3. Now from one the sides, cut 2 more times.

In 6 cuts you get 27 pieces. Also equal piece cubes if properly cut. But not sure if this is minimal cuts you can achieve, unless there is a mathematical explanation from someone here.

NOTE: <<< Incorrect code >>> No time to fix it, got to run, something like below will work.

int main() {

int N = 27; // target cuts // you can change this

int dimensions = 3; // cube 3D object

for (int i=0; if Tom McIntyre

M.S. in Physics, University of California, Irvine (Graduated 1996)Author has 51 answers and 152.2K answer views4y

Guys, does anyone know the answer?