Is it possible to colour the faces of $27$ unit cubes and arrange them to form $3\times3\times3$ cube with all exterior faces red,then white,then blue Is it possible to colour the faces of $27$ unit cubes so that one can arrange them to form $3\times 3\times 3$ cube with all exterior faces red, then rearrange to form a cube with all exterior faces white, then blue?
There have be at least $24$ or $23$ cubes with all three of their faces in one colour... Thereafter trying various possibilities my guess seems to be no,  but I'm not sure and don't know how to prove it.

This is possible. 
Let $(x,y,z)$ represent a cube with $x$ red faces, $y$ white faces, and $z$ blue faces. Then one way to paint the small cubes is as follows:


*

*Make one each of $(3,3,0)$, $(3,0,3)$, and $(0,3,3)$.

*Make three each of $(1,2,3)$, $(1,3,2)$, $(2,1,3)$, $(2,3,1)$, $(3,1,2)$, $(3,2,1)$.

*Make six of $(2,2,2)$.


You can check that for each color, there is $1$ cube with no faces of that color, $6$ cubes with one face of that color, $12$ cubes with two faces of that color, and $8$ cubes with three faces of that color.
