The entire exterior of a large wooden cube is painted red

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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 ...

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.

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.

Just for purpose of simplification, this formula can be used

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

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

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

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

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

Hi Guys,
When I try to substitute n = 4, it seems that both B and C works. Please help!

For plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

A nice example of plug in method usage.

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.

Am I missing something here? 54 - 36 + 8 = 10 (not 26)

54 - 36 + 8 = 26, not 10.

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

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

a formula? What's the logic/reasoning behind it? How does one derive this during the exam!

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.

Im confused about this topic, could someone please help me explain the logic behind, why there's only 1 cube in the middle that's unpainted?

If we look at a rubik's cube ... theres 6 faces .... 9 "smaller cubes" on the face of the cube = 54 smaller cubes

Or on another logic;

3^3 = 27 smaller cubes ----> Rubik's cube has 6 faces , therefore 27/6 = 24 (24 "smaller cubes" on the face of the rubik's cube) and then 3 remaining on the "inside".

Please help me get this logic right in simple language as possible! Bunuel or anybody else.

Thank you!

how did you get the formula for: Number of cube inside = \((n-2)^{3}\)?
- something you need to memorize?
- i can't even visualize it...a cube has 6 faces (top/bottom, +4 around). if we assume a particular cube has 3 tiny cubes per row & column, we can conclude there will be an "innermost" tiny cube that need not be painted red (it would be buried inside the cube). how do we know that in order to find this innermost cube, the formula is: \((n-2)^{3}\)

more in-depth analysis would be much appreciated

For this problem, pick the smallest n that will satisfy the problem. In this case, n =3 satisfies the criterion. You can then draw a cube with 27 individual pieces. Only the middle cube will be unpainted. So you want to then backsolve from the answer choices until you find 26.

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:

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

I did as Bunuel, but it took 8 minutes to come to a correct answer, hope more practice will help me reducing time