A cube is sectioned by $33$ of planes parallel with faces of cube in many parallelipipeds. If the number of parallelipipeds with faces that coincide with the faces of cubes is half of the total number. Find the total number.
My idea: Let $a, b, c $ the number of the three types of planes.
Then there are $(a+1)(b+1)(c+1) $ parallelipipeds.
Also, there are $(a-1)(b-1)(c-1) $ parallelipipeds that hasn't the same faces as cube.
So
$(a+1)(b+1)(c+1)=2 (a-1)(b-1)(c-1) $
Now I am stuck.
If you set $x=a-1$, $y=b-1$, $z=c-1$, your equation can be rewritten as $$ (x+2)(y+2)(z+2)=2xyz $$ and substituting here $z=30-x-y$ we get $$ 128+60(x+y)-2(x^2+y^2)-32xy+(x+y)xy=0. $$ Make now the change of variables: $u=xy$, $v=x+y$: the above equation becomes $$ 128-28u+60v+uv-2v^2=0, $$ which can be solved for $u$ to give $$ u=2{64+30v-v^2\over28-v}. $$ It is now tedious but straightforward to check all the possible integer solutions, for $0\le v\le27$, which in turn give integer values of $x$, $y$, $z$. It turns out that there are two possible solutions for $(x,y,z)$: $$ (4,12,14),\quad\hbox{or}\quad(5,7,18). $$