# Sunday Teaser 2648 – Painted Cubes

### by Graham Smithers

I found some painted cubes whose sides were varying whole numbers of centimetres.

After choosing some of these cubes whose edges were consecutive integers, I cut them all into smaller cubes with edge lengths of one centimetre.

I found that half of the resulting cubes had no painted faces.

How many of the small cubes had no painted faces?

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An alternative solution requires a bit of analysis. If we let $$m$$ be the edge size of the smallest cube in the sequence, and $$n$$ be the number of cubes, then the number of unit cubes is given by $C(p,n)=(np/4)(4p^2+n^2-1)$ where $$p=m+(n-1)/2$$, the average side length of the cubes.
For the inner cubes the side lengths are 2 smaller and we are told that the number of cubes with no painted faces is one half of the total number – that is $$C(p,n)=2C(p-2, n)$$. This simplifies to give:$n^2= -\frac{(4p^3-48p^2+95p-60)}{(p-4)}$ Here we can solve for $$n$$ and $$p$$ by stepping $$p$$ in half integers greater than 4. So if we let $$q=2p$$ we obtain $n^2= -\frac{(q^3-24q^2+95q-120)}{(q-8)}$ which is the form used in the program below.