        \problemdata       % Takes 5 *mandatory* arguments
        {cube-split}             % latex-friendly label for the prob.
        {induction}  % The topic(s) of the problem content.
        {Rosen, Exercise 3.2.63}               % Source (if known)
        {F01 tut2-2} % Usage (list ones you are aware of).
        {}        % Last revision info (author, date).

\begin{problem}
Rosen, Exercise 3.2.63

\solution{%page 96 (63)

This problem is very similar to Example $12$; the only difficulty is in
visualizing what's happening in three dimensions.  The basis step $(n=1)$
is trivial, since one tile coincides with the solid to be tiled.  To make
this read a little easier, let us call a $1 \times 1 \times 1$ cube a
``cubie,'' and let us call the object we are tiling with, namely the $2
\times 2 \times 2$ cube with one cubie removed, a tile.  For the inductive
step, assume the inductive hypothesis, that the $2^n \times 2^n \times
2^n$ cube with one cubie removed can be covered with tiles, and suppose
that a $2^{n+1} \times 2^{n+1} \times 2^{n+1}$ cube with one cubie removed
is given.  We must show how to cover it with tiles.  Think of this large
object as split into eight octants, by planes parallel to the faces.  The
missing cubie occurs in one of these octants.  Now position one tile with
its center at the center of the large object, so that the missing cubie in
the tile lies in the octant in which the large object is missing its
cubie. This creates eight $2^n \times 2^n \times 2^n$ cubes, each missing
exactly one cubie-one in each octant.  By the inductive hypothesis, we can
fill each of these smaller objects with tiles.  Putting these tilings
together gives us the desired tiling of the $2^{n+1} \times 2^{n+1} \times
2^{n+1}$ cube with one cubie removed, as desired.}

\end{problem}
