Waste less time on Facebook — follow Brilliant.


Recently, I was playing with my two year old brother. There was an old \(3\times 3\) Rubik's cube lying around and my dad posed a simple problem:

How many selections of \(2\times 2\) cubes can be made from a \(3\times 3\) Rubik's cube?

The answer to the above question is \(8\) and it is so because each corner cube is part of a unique \(2\times 2\) selection.

Now, how many selections of \(1 \times 1\) cubes can be made from a \(3\times 3\)? The answer here is too simple and is \(27\).

And the number of selections of a \(3\times 3\) cube from a \(3\times 3\) is \(1\).

\(1, 8, 27\).... Reminded me of something. This note is an attempt to prove or disprove the following hypothesis.

In an \(n \times n\) Rubik's cube, the number of selections of solid cubes of edge length \(m(\le n)\) is \((n-m+1)^3\)

I am convinced that there is an elegant solution to this problem and that the hypothesis is true.

Please post your ideas. Thanks!

Note by Raghav Vaidyanathan
2 years ago

No vote yet
1 vote


Sort by:

Top Newest

Consider one dimension first. Given an array of length \(n\), we can place a smaller array of length \(m\) in \(n-m+1\) positions. This is the same for all three dimensions, bringing the total number of placings to be \((n+m-1)^3\). Daniel Liu · 2 years ago

Log in to reply

@Daniel Liu Great explanation! Thanks! Raghav Vaidyanathan · 2 years ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...