Determine whether it is possible to find a cube and a plane such that the distances from the vertices of the cube to the plane are 0,1,2,3,....,7 Problem: Determine whether it is possible to find a cube and a plane such that the distances from the vertices of the cube to the plane are $0,1,2,3,....,7.$
Now in the solution, the author states that: 

If we consider a cube of edge length $1$ unit with the coordinates
  $(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0)$ and
  $(1,1,1)$ we observe that these are binary representations of the
  numbers $0,1,2,3...,7.$ This motivates us to consider the plane
  $x+2y+4z=0.$

I didn't get how the observation of coordinates being binary equivalents of the numbers $0,1,2..,7$ motivate the author to consider the plane $x+2y+4z=0.$ Please explain.
 A: For a plane with equation
$$P \equiv ax + by +cz +d=0$$ the distance of a point with coordinates $(x_0,y_0,z_0)$ to the plane is
$$D=\frac{\vert ax_0 + by_0 +cz_0 +d \vert}{\sqrt{a^2+b^2+c^2}}$$
If you take $a=1, b= 2, c= 4, d=0$, then as explained by the author the quantity $D$ will take all integer values in $\{0, \dots, 7\}$ Hence the distance of the unit cube vertices to the plane will be in $\{0 /\sqrt{21}, \dots, 7 /\sqrt{21}\}$ as $21= 1^2 + 2^2 +4^2$.
To get the solution you're looking for, transform the unit cube with an homothety centered at the origin and with ratio $\sqrt{21}$. You'll verify that the plane $$P \equiv x+2y+4z$$ and transformed unit cube are answering the initial question. 
A: Consider a binary number $\overline{x_n x_{n-1}\ldots x_1x_0}$. Its value is equal to $$1\cdot x_0 + 2\cdot x_1 + 4 \cdot x_2+\cdots +2^{n-1}x_{n-1} + 2^nx_n,$$
hence in the case $n=2$ it indeed seems logical to start by considering a plane whose coefficients correspond to the decomposition of values of a binary number.
