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.
http://en.wikipedia.org/wiki/De_Bruijn%27s_theorem
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.