No of unit cubes in a nnn cube

Question

Imagine a cubic array made up of an $n\times n\times n$ arrangement of unit cubes: the cubic array is n cubes wide, n cubes high and n cubes deep. A special case is a $3\times3\times3$ Rubik’s cube, which you may be familiar with. How many unit cubes are there on the surface of the $n\times n\times n$ cubic array?

As far as I can see there are 27 unit cubes in a $n\times n\times n$ rubik cube. But the answer says something different. There are total $6n^2$ squares are present in $n\times n\times n$ cube. But after that I cant proceed.

Please help :)

You mean in an $n \times n \times n$ cube composite, how many cubes are there which you can see from the outside? — Sarvesh Ravichandran Iyer, Apr 23 '16 at 04:39
In the $3\times 3\times 3$ there are $26$, all but the central cubelet. More generally there are $n^3-(n-2)^3$, which can be written in other less informative ways. — André Nicolas, Apr 23 '16 at 04:41
@AndréNicolas could you please explain a bit further .. the logic why $ (n-2)^3 $ is subtracted — ViX28, Apr 23 '16 at 04:42
In the $n\times n\times n$ case, the invisible part is $(n-2)\times (n-2)\times (n-2)$, so the visible part is all $n^3$ of them except for the (n-2)^3$ invisibles. — André Nicolas, Apr 23 '16 at 04:45
@AndréNicolas every time you answer I feel like my question got half solved :) thanks :) — ViX28, Apr 23 '16 at 04:54

Stella Biderman · Accepted Answer · 2016-04-23T04:51:18.400

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So, i think that thing you're missing is that you're counting the number of squares on the surface; not the number of cubes. For example, a corner piece of a Rubix cube is one cube but contributes three squares.

In a $n\times n\times n$ cube, you have the outer layer of cubes counting, and the ones on the inside not counting. The easiest way to see this is recursive: for a $n\times n\times n$ cube, you have an $(n-2)\times (n-2)\times (n-2)$ cube on the inside that is coated by a number of cubes that form the cubes on the outside. Subtracting $$n^3-(n-2)^3=6n^2-12n+8$$ gives the final answer.

This generalizes immediately to higher dimensions... In dimension $k$ the answer is $$n^k-(n-2)^k$$

edited Apr 23 '16 at 04:51

answered Apr 23 '16 at 04:44

Stella Biderman

31,155

that will be $12n$ instead of $6n$. – ViX28 Apr 23 '16 at 04:50
@ViX28 thanks! Fixed that and added the general formula – Stella Biderman Apr 23 '16 at 04:51

score 1 · Answer 2 · answered Apr 23 '16 at 04:52

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There are $6$ faces with $n^2$ cubes on each face for a total of $6n^2$ cubes. The eight cubes on the vertices are counted $3$ times each so we must subtract $16$ to get $6n^2-16$ cubes. Likewise, there are $12$ edges, each with $n-2$ cubes that have been double counted so we must subtract $12(n-2)$ to get $6n^2-16-12n+24=6n^2-12n+8$ cubes.

answered Apr 23 '16 at 04:52

John Douma

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More simply: $6n^2$ cubes on each face. We have counted the edges too many times, so subtract $12n$. But now we have undercounted the vertices, so add $8$ again. This gives $6n^2-12n+8$. – Théophile Apr 23 '16 at 05:03

No of unit cubes in a n*n*n cube

2 Answers2

No of unit cubes in a nnn cube