A hundred identical cubic boxes are currently arranged in four cubes:

Question

A hundred identical cubic boxes are currently arranged in four cubes: a single cubic box, a 2 x 2 x 2 cube, a 3 x 3 x 3 cube, and a 4 x 4 x 4 cube. These four are not touching each other. All outward faces are painted and all inward faces are not painted. These four cubes are going to be dismantled and reassembled as a flat 10 x 10 square. The top and all the edges of this 10 x 10 square must be painted, but there is no requirement for paint on the bottom. How many individual faces will have to be painted to accommodate the requirements of this new design?

(A) 0
(B) 5
(C) 9
(D) 16
(E) 27

This is one of a set of 15 challenging GMAT Quant practice problems. To see the whole collections, as well as the OE for this question, see:
Challenging GMAT Math Practice Questions

Mike :-)

arshak90 · Accepted Answer

The answer to this problem can be obtained by simply counting the small cubes that are not painted outward. 
In single cubic box - no such cubes
In 2*2*2 cubic box - also no such cubes
In 3*3*3 cubic box - 1 such cube that is located in the centre of the cubic box
In 4*4*4 cubic box - 8 such cubes that are also located in the centre of the cubic box
As we have to use all 100 small cubes to make 10*10 square, thus 8+1=9 faces of 9 not painted cubes have to be painted.

utkarshthapak · Answer

Does this difficulty question come in GMAT?

Tj54 · Answer

The challenge here appears to me that when I'll dismantle all the little cubes and reconstruct a new 10x10 square (actually not a square, rather a rectangular solid of 10x10x1) then some previously painted faces are going to be lost. How would I determine those total lost painted surfaces?

arunavamunshi1988 · Answer

I could not solve this, despite drawing a series of figures. Need an OE...

Bunuel · Answer

Have you read the highlighted part?

mondirachakraborty · Answer

Bunuel, will you please provide us with an easy solution? I have gone through the OE part but wasn't able to understand.

gmatexam439 · Answer

My approach was similar to what has been explained by Mike and took me around 2 and a half minutes to solve. But Please clarify do such questions come in GMAT? Its pure thought process, breaking up a 10x10 into different pieces like a matrix movie, i mean pure imagination and no mathematical concept used.

sjavvadi · Answer

The question is basically, when we lay the required 10x10x1 structure, how many additional faces on the top do we have to paint.

From the 4 cubes initially set up, even if we have 1 face that is painted we can place that cube in the final structure such that the painted face is on top.

So from the 1x1x1 cube: We have 1 cube used and and 1 cube where at least 1 side is painted = 1
So from the 2x2x2 cube: We have 8 cubes used and and 8 cubes where at least 1 side is painted =8
So from the 3x3x3 cube: We have 27 cubes used and and 26 cubes where at least 1 side is painted (the middle cube does not have even 1 side painted) =26
So from the 4x4x4 cube: We have 64 cubes used and and 56 cubes where at least 1 side is painted (the middle 8 cubes do not have even 1 side painted) =56.

Therefore, total where we have atleast 1 face painted = 1+8+26+56 = 91.

So an additional 9 cubes need to be painted so that all the faces on the top are painted.

mikemcgarry · Answer

Dear mondirachakraborty & gmatexam439 , I'm happy to respond. :-)

First of all, mondirachakraborty , this is a HARD question. Some questions on the GMAT look hard but there's some simplifying trick. For this question, there's no easy trick: the question is quick only if you have the spatial/visual intuition to think through the ideas in the OE quickly. My friend, if you are asking for some way for the hard to be made easy for you, you are asking the wrong question. The question is how to rise to the challenge of what you now find hard, so that eventually it can seem easy to you. If you find an OE that you don't understand, don't look for a way to evade it. Instead, dig deeper into anything you don't understand. Dissect it line by line. If there are particular points that you don't understand. ask me (the author of this question) about those very specific questions. Education is not something experts such as Bunuel and I do to you: instead, education is a process you do to yourself, by yourself, and for yourself. It depends primarily on your dedication, diligence, and engagement. Go all in, and we experts will do whatever we can to support you. And, gmatexam439 , this is a hard question, probably about at about the upper limit of what the CAT would throw at a student who is acing everything else in the Quant section. In the OE, i spelled out everything in meticulous detail, but if you have good spatial intuition, you might be able to do this in under 20 seconds. We need four corners--more than enough corners. We need 32 edges--more than enough edges and left over corners, so we are covered. For everything else, we just need one side painted, so the only ones that need paint are the unpainted "inner cubes," the 1 inside the 3x3x3 and the 8 inside the 4x4x4. (Any cube of nxnxn has an inner cube of (n-2)x(n-2)x(n-2) that doesn't see the light of day.) Paint 9 faces and we are all set. As with many challenging Quant questions, it's simply a matter of mentally dissecting the scenario in the right way: when you do that, the answer simply appears. Does all this make sense? Mike :-)

sriramsundaram91 · Answer

First of all, if the question is understood then half battle is won.
Let me explain the question first there are 100 identical cubes, i.e 1*1*1=1 cube, 2*2*2= 8 (1*1*1) cubes, 3*3*3 = 27(1*1*1) cubes and 4*4*4 = 64(1*1*1) cubes.

The best way to solve the problem is buy a 2*2,3*3,4*4 rubiks cube or find images of such on the internet.

Now if you hold a 1*1*1 cube, the cube has six faces, so paint all of the faces.(Paint the faces you can see is what that outward inward thing is).
Now in a 2*2*2 cube you can have 8 1*1*1 cubes, 4 facing you and 4 behind the 4 facing you. so you could technically paint the faces of all the 8 cubes(don't care about the face not painted unless you could not paint the whole cube).

Now in a 3*3*3 ,if you could find an image of that in the internet you can find that one cube of all 27 would remain unpainted (1 unpainted cube)

Similarly if you think for long enough you can find 8 unpainted 1*1*1 cubes from 64 (1*1*1) cubes of 4*4*4 box. So in total we need(1+8)= 9 cubes to be painted. (Why because we need to arrange all 100 cubes in 10*10 matrix (Note not 10*10*10) 2D just layout 100 cubes in 10*10 in a desk.)

vishwash · Answer

yeah for this level question... i suppose.. u should get 99% questions correct in the cat and then the cat stumps u with this one type or the other questions present in the highlighted post

ScottTargetTestPrep · Answer

First let’s determine the number of cubes that are already painted with at least one face:

The single cube, all 8 of the 2 x 2 x 2 cube, 26 of the 3 x 3 x 3 cube (since only 1 interior cube is not painted), and 56 of the 4 x 4 x 4 cube (since only 2^3 = 8 interior cubes are not painted).

Therefore, so far we have 1 + 8 + 26 + 56 = 91 cubes are painted with at least one face. So we need 9 more cubes to be painted so that all 100 cubes will be painted with at least one face.

Next, we need to make sure that we have enough cubes with two and three faces painted. Note that when the cubes are reassembled as a flat 10 x 10 square, the four corner cubes must have three faces painted. Also, of the remaining 10 - 2 = 8 cubes on each edge, 8 x 4 = 24 cubes must have two faces painted. The single cube together with three cubes from the 2 x 2 x 2 cube have at least three faces painted; thus these cubes can be used on the corners of the flat 10 x 10 square. We have 8 - 3 = 5 cubes with three faces painted remaining from the 2 x 2 x 2 cube. On the 3 x 3 x 3 cube, we have 26 cubes painted and only 6 of these cubes have only one side painted (the cubes that are located at the center of each face); therefore 26 - 6 = 20 cubes on the 3 x 3 x 3 cube have at least 2 sides painted. Together with the 5 cubes from the 2 x 2 x 2 cube, we have more than 24 cubes with at least 2 sides painted; therefore we only need the 9 faces that we determined earlier to be painted.
 
Answer: C
.

smlprkh · Answer

Newly Assembled Cube - 100 Total Cubes (4 need 3 sides painted, 32 need 2 sides painted, 64 need 1 side painted)

1X1X1 cube has 1 cube with 6 sides painted
2X2X2 cube has 8 cubes with 3 sides painted
3X3X3 cube has 8 cubes with 3 sides painted, 12 cubes with 2 sides painted and 6 cubes with 1 side painted
4X4X4 cube has 8 cubes with 3 sides painted, 24 cubes with 2 sides painted and 24 cubes with 1 side painted

This leaves an excess of 9 cubes with 1 side painted required. C

Fdambro294 · Answer

“Individual” faces refers to each of the faces of 1 x 1 x 1 cubes that make up each of the 4 interim cubes.

It is not at all made clear at first that this is what is meant by “individual” faces. One could read the question and believe that the “individual” faces referred to each of the faces of the interim cubes (I.e., 6 faces on the 4x4x4 cube ; 6 faces on the 3x3x3 cube ; etc.)

As such, we need to use all 100 1x1x1 cubes that make up each interim cube in our 10x10 “flat” square (which is really a 3-D figure with dimensions 10 x 10 x 1)

Step 1: paint all the faces of the interim cubes and determine how many faces we have painted

1x1x1 —- 1 cube

1 cube with 6 faces painted

2x2x2 —— 8 cubes

8 vertices: so 8 cubes will have 3 faces painted

This accounts for all the cubes

3x3x3 —— 27 cubes

8 vertices: 8 cubes will have 3 faces painted

12 edges: 1 cube on each edge with 2 faces painted =

12 cubes with 2 faces painted

6 faces: on each face there is 1 cube with only 1 face painted =

6 cubes with 1 face painted

1 cube Not painted

4x4x4 ——— 64 cubes

8 vertices: 8 cubes with 3 faces painted

12 edges: 2 on each edge will have 2 faces painted =

24 cubes with 2 faces painted

6 faces: 4 cubes on each face =

24 cubes with only 1 face painted

8 cubes unpainted

Step 2: how many individual faces (1x1x1 faces) do we need to paint the top and edges of the 10x10x1 square

We will be using all 100 individual 1x1x1 squares and have to determine how many more 1x1x1 faces need to be painted.

4 vertices: need 4 cubes with 3 faces painted

4 edges: 8 cubes along each edge that will have 2 faces painted

Need 32 cubes with 2 faces painted

Middle of top face of square: (8) (8) = 64 faces showing with just 1 face painted

(3rd) using each one of the cubes to make the “flat” square:

First, we can use the 8 cubes with 3 faces painted from the 4x4x4

Place 4 of these on the corners of the new “flat” square.

We can use the 4 left over on the edges. That leaves 32 - 4 = 28 needed with 2 faces

We can then take the 24 cubes with 2 faces painted from the 4x4x4 cube and use those. Then take 4 of the 2 face cubes from the 3x3x3 cube (leaves us with 8 left with 2 faces from the 3x3x3 cube)

Then use the 24 cubes with 1 face painted to help fill in the middle 64 faces of this big flat square.

At this point, all we have left is 64 - 24 =

—40 faces showing in the middle of the flat square that need to be painted.

—What we have remaining with faces painted

1 of the 1x1x1 cubes

3x3x3 cube: 
—8 cubes with 3 faces
— 8 cubes with 2 faces 
—6 cubes with 1 face

2x2x cube:

— 8 cubes with 2 faces

We can only arrange these cubes such that one of their painted faces is showing.

That is: 1 + 8 + 8 + 6 + 8 = 31 cube we can use to fill in the middle of this big flat 10x10x1 square.

We needed 40 though.

40 - 31 = 9 of the 1x1x1 faces remain that need to be painted

Answer is 9

Posted from my mobile device

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A hundred identical cubic boxes are currently arranged in four cubes:

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