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Smaller cubes that have no paint?

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by gmattesttaker2 » Sat Feb 15, 2014 4:57 pm
Hello,

Can you please assist with this:

A wooden cube of side 6 is painted blue; the cube is then divided into 216 equal
cubes, how many of the smaller cubes have no side painted blue?

(A) 124
(B) 108
(C) 96
(D) 64
(E) 48

OA: D


Thanks a lot,
Sri
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by Patrick_GMATFix » Sat Feb 15, 2014 5:21 pm
What a great question Sri.

(cubes-along-length) * (cubes-along-width) * (cubes-along-height) = 216 total cubes. Since length,width and heights are the same, we can say that (cubes-along-any-edge) = cubic root of 216 = 6.

All the cubes that make up the skin of the big cube are painted. The unpainted cubes are the cubes on the inside. Imagine the big cube as a blue cake with six layers (top to bottom). The top and bottom layers will have only painted cubes. The 4 middle layers will be identical, such that only the 4 by 4 inner cubes are unpainted. So there are 16 unpainted cubes in each of these 4 middle layers. there are 16 * 4 = 64 unpainted cubes.

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by ankitbagla » Sat Feb 15, 2014 6:27 pm
Good visual question
let us consider that larger cube volume is 216 units. now each smaller cube size will be 1 X 1 X 1 . Now the outer layers are painted with blue. So all the cubes in the outer layer will have one side painted . So their numbers need to be deducted. by visualizing you can easily see that what will be left inside only a 4 X 4 X 4 cubes by removing the larger of unit from each faces . So the left over will be 64 cubes .
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by GMATGuruNY » Sun Feb 16, 2014 12:13 am
gmattesttaker2 wrote:Hello,

Can you please assist with this:

A wooden cube of side 6 is painted blue; the cube is then divided into 216 equal
cubes, how many of the smaller cubes have no side painted blue?

(A) 124
(B) 108
(C) 96
(D) 64
(E) 48
The larger cube is 6*6*6 = 216 -- the cube of an integer.
When the outer layer of the cube is painted blue and removed, what remains is a SMALLER CUBE that also must be the cube of an integer.
Only one answer choice is the cube of an integer:
64 = 4*4*4.

The correct answer is D.
Last edited by GMATGuruNY on Sun Feb 16, 2014 8:47 pm, edited 1 time in total.
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by [email protected] » Sun Feb 16, 2014 8:21 pm
Hi Sri,

Each of the solutions offered by Mitch, ankitbagla and Patrick are correct; that approach relies on your ability to "remove" the "outer cube" (made up of all the little cubes that are partially painted) and focus on the "inner cube" (all the little cubes that have no paint on them). It's an elegant approach. If you can't "see" that approach though, there is another way to get to the correct answer.

We're told that the large cube is broken up into 216 smaller cubes. If you take one "face" of that big cube (we'll call it the "front" face), you'll have a 6x6 set of little cubes that all have paint on them. On the opposite side of the large cube (the "back" face) is another 6x6 set of little cubes that all have paint on them.

Front + Back = 36 + 36 = 72 cubes with paint on them.

Now look at the "left" face of the large cube. It already has 12 painted little cubes (6 from the "front" face and 6 from the "back" face), so we're not going to count those again. That leaves 36 - 12 = 24 new faces with paint on them. The same number will appear on the "right" face of the large cube.

Left + Right = 24 + 24 = 48 additional cubes with paint on them.

Now look at the "top" face of the large cube. It already has 20 painted little cubist (6 from the "front" face, 6 from the "back" face, 4 additional from the "left" face and 4 additional from the "right" face). That leaves 36 - 20 = 16 new faces with paint on them. The same number will appear on the "bottom" face of the large cube.

16 + 16 = 32 additional cubes with paint on them.

72 + 48 + 32 = 152 little cubes with paint on them.

216 total - 152 with paint = 64 with NO paint.

Final Answer: D

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