Is it possible to fill a box with dimensions $10\times10\times10$ using bricks of dimensions $1\times1\times4$?

If yes, how?

I think De Bruijn's theorem on harmonic bricks could be helpful, but I don't know if this theorem can be applied when 2 edges have length 1. I am not sure that the brick can be considered as harmonic. More info on De_Bruijn's theorem can be found here.


I don't have aceess to any paper where the theorem is explained in more detail.


No, it is not. This can be seen via generating functions. We can position the box in 3-space with one corner at the origin so that it occupies $(a,b,c)$ for $0\leq a,b,c \leq 9$. Let us associate each point $(a,b,c)$ with the polynomial $x^ay^bz^c$. The sum of the points in the box can then be expressed as:

$$\mathrm{BOX}(x,y,z) = \sum_{a=0}^9 \sum_{b=0}^9 \sum_{c=0}^9 x^ay^bz^c$$

On the other hand, each brick can be expressed as either $x^ay^bz^c(1+x+x^2+x^3)$, $x^ay^bz^c(1+y+y^2+y^3)$, or $x^ay^bz^c(1+z+z^2+z^3)$ for some integers $a,b,c$, depending on the brick's orientation. The sum of the points in the union of the bricks can be expressed as:

$$\mathrm{BRICKS}(x,y,z) = F(x,y,z)(1+x+x^2+x^3) + G(x,y,z)(1+y+y^2+y^3) + H(x,y,z)(1+z+z^2+z^3)$$

for some polynomials $F, G, H$.

But $\mathrm{BOX}(i,i,i) = -2+2i$ while $\mathrm{BRICKS}(i,i,i)=0$, implying that the two polynomials are not identical, and thus that the desired partitioning is impossible.

  • $\begingroup$ Beautiful solution! Is it possible that $\mathrm{BOX}(i,i,i)=(i-1)^3=2+2i$? $\endgroup$ – wnvl Oct 5 '14 at 21:50
  • $\begingroup$ I just found out that it is $(i+1)^3=-2+2i$. Please, ignore my previous remark. $\endgroup$ – wnvl Oct 5 '14 at 22:14
  • 1
    $\begingroup$ @wnvl Thanks! This is a technique I learned for this type of problem back in my olympiad days. It can be used to solve a variety of tiling problems in any number of dimensions. $\endgroup$ – dshin Oct 5 '14 at 23:35

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