The entire exterior of a large wooden cube is painted red

Question

The entire exterior of a large wooden cube is painted red, and then the cube is sliced into n^3 smaller cubes (where n > 2). Each of the smaller cubes is identical. In terms of n, how many of these smaller cubes have been painted red on at least one of their faces?

A. 6n^2
B. 6n^2 – 12n + 8
C. 6n^2 – 16n + 24
D. 4n^2
E. 24n – 24

A. 6n^2
B. 6n^2 – 12n + 8
C. 6n^2 – 16n + 24
D. 4n^2
E. 24n – 24

Bunuel · Accepted Answer

Say n=3. So, we would have that the large cube is cut into 3^3=27 smaller cubes: Red Cube.png Out of them only the central little cube won't be painted red at all and the remaining 26 will have at least one red face. Now, plug n=3 and see which one of the options will yield 26. Only B works: 6n^2 – 12n + 8 = 54 - 36 + 8 = 26. Answer: B.

kundankshrivastava · Answer

Number of cube inside = (n-2)^3

(n-2)^3 cubes have no colored faces.

Remaining cubes will have at least one face colored red.

Remaining cubes = n^3 -(n-2)^3

= n^3 -( n^3 - 8 - 3*n*2(n-2))
=n^3 - (n^3 - 8 - 6n^2 +12n)
=6n^2-12 n +8

Answer is B

Bunuel · Answer

Similar questions to practice: if-a-4-cm-cube-is-cut-into-1-cm-cubes-then-what-is-the-107843.html a-big-cube-is-formed-by-rearranging-the-160-coloured-and-99424.html a-large-cube-consists-of-125-identical-small-cubes-how-110256.html 64-small-identical-cubes-are-used-to-form-a-large-cube-151009.html Hope it helps.

abhijeetjha · Answer

Obviously bunuel's solution is mindblowing and the best approach we need in such PS questions ...But just while brainstorming trying to find the solution to this problem i reached here ...Try to visualise that all the smaller cubes which lie on the exterior face of the larger wooden cube have one or more faces painted red...rest all other cubes which lie beneath the first layer of cube wont have any faces painted red...

Now if we can visualise the situation .....we can see that if we remove the external layers of cube ..we will be left with cubes having none of their faces red coloured...and if we remove this external layers of cube we are basically removing one cube from each side symmetrically ...so we will be left with a cube having dimensions of (n-2).... so basically we will be left with (n-2)^3 cubes ... now if we want to find the number of cubes in the larger cube having one or more faces as red we can deduct (n-2)^3 from n^3...

so number of cubes painted red =n^3 - (n-2)^3
 =n^3-(n^3-8-6n^2+12n)= 6n^2-12n+8.......B

this solution is fairly easy too just need little bit of visualization........
 Bunuel please correct me if i am wrong somewhere ...

davidfrank · Answer

Hi Bunuel,

As usual great explanation. I just have one question , what if we choose n to be 4 or 5. The visualization then becomes little difficult. What would then a better approach solve such question.

Bunuel · Answer

You could apply the same logic:

Say n=5. So, we would have that the large cube is cut into 5^3=125 smaller cubes. Out of them (5-2)^3=27 little cubes won't be painted red at all and the remaining 125-27=98 will have at least one red face. Now, plug n=5 and see which one of the options will yield 98.

But you can use n directly: 
Total = n^3
Not painted: (n-2)^3

Difference = n^3 - (n-2)^3 = 6n^2 – 12n + 8.

Hope it helps.

PareshGmat · Answer

Just for purpose of simplification, this formula can be used

a^3 - b^3 = (a-b) (a^2 + ab + b^2)

= (n - n + 2) (n^2 + n(n-2) + (n-2)^2)

= 2 (n^2 + n^2 - 2n + n^2 - 4n + 4)

= 2 (3n^2 - 6n + 4)

= 6n^2 - 12n + 8

youneverknow · Answer

The (N-2) approach is right. However 2 is again, an assumption. If I take a side x, the the volume of inner cube becomes  (N-2x)^3  .

Here's how:

Assuming volume of larger cube = N^3 
Volume of smaller cubes formed =  N^3/n^3  or (N/n)^3

Thus side of smaller cubes = N/n =  x (say)
Therefore, volume of smaller cube = (N-2x)^3

Notice that all the squares formed in this cube won't be touched by the paint.

Therefore, total volume of cubes that will have a face painted =  N^3-(N-2x)^3 
And no. of cubes with this volume can be found by dividing the above equation by  x^3 .

If we start-  (N^3-(N-2x)^3)/x^3

=  (8x^3 + blah blah..)/x^3 
= 8 + something

At this point, we can notice that only one option consists 8 as a constant. Otherwise, if you choose solve it, you get the same result.

EMPOWERgmatRichC · Answer

Hi All,

This question can be solved by TESTing VALUES. Let's TEST N = 3 (if you think about a standard Rubik's cube, then that might help you to visualize what the cube would look like).

So now every "outside" face of the Rubik's cube has been painted. There's only 1 smaller cube of the 27 smaller cubes that does not have paint on it (the one that's in the exact middle). Thus, 26 is the answer to the question when we TEST N=3. There's only one answer that matches....

Final Answer:  B

GMAT assassins aren't born, they're made, 
Rich

ScottTargetTestPrep · Answer

The number of smaller cubes that have no faces painted is (n - 2)^3. Therefore, the number of smaller cubes that have at least one face painted is:

n^3 - (n - 2)^3 = n^3 - (n^3 - 6n^2 + 12n - 8) = 6n^2 - 12n + 8

Alternate Solution:

Let n = 3. Then we know we have 3^3 = 27 smaller cubes, and 26 of them (all except the innermost cube) will have at least one face that is painted red.

If we plug in 3 for n in each answer choice, we see that choice B is the only one that gives 26 as the answer: 6 * 9 - 12 * 3 + 8 = 54 - 36 + 8 = 26.

Answer: B

mangamma · Answer

problem never specifies a real number for any of the steps, so select a smart number for n, say n = 3. In this case, the original cube is painted and then cut into a 3 × 3 × 3 assemblage of 27 smaller cubes. Imagine the top face of the original (large) cube (if you know what a Rubik’s Cube is, picture one!). Every cube on that face has been painted on at least one side. The same is true for the bottom face. Now think about the middle “slice” of the cube. This “slice” contains a total of 9 cubes, but only the one in the very middle has not been painted on any of its sides.

Therefore, only one of those 27 cubes—the one in the middle of the structure—remains unpainted. When n = 3, the answer is 26.
 
Plug n = 3 into the answers and look for 26:
 
(A) 6(32) = 6(9) = 54
(B) 6(32) – 12(3) + 8 = 54 – 36 + 8 = 26
(C) 6(32) – 16(3) + 24 = 54 – 48 + 24 = 30
(D) 4(32) = 4(9) = 36
(E) 24(3) – 24 = 72 – 24 = 48

bumpbot · Answer

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The entire exterior of a large wooden cube is painted red

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