Faustus wrote: The faces of a box with integer edge lengths are painted green. The box is partitioned into unit cubes. Find the dimensions of the box if the number of unit cubes with no green face is one third of the total number of cubes. note , that if the box with dimensions $ a \times b \times c $ is painted green on all side's then number of cubes which are left unpainted will be $(a-2)(b-2)(c-2)$ As given, $(a-2)(b-2)(c-2)=\frac{abc}{3}$ $3 = \frac{abc}{(a-2)(b-2)(c-2)}$ now if all the dimensions are $\ge 7 $ then $3= \frac{abc}{(a-2)(b-2)(c-2)} \le \frac{343}{125} \simeq 2.74$ this means that at least one of the dimension is $\le 7 $ Contradiction Let suppose $c \le 7$ , then possible value's for $c$ are $3,4,5,6$ now do trivial case chase and you will get the possible dimensions as $ 7 \times 30 \times 4 \\ 8 \times 18 \times 4 \\ 9 \times 14 \times 4 \\ 10 \times 12 \times 4$ $ 5 \times 27 \times 5 \\ 6 \times 12 \times 5 \\ 7 \times 9 \times 5 $ $6 \times 8 \times 6 $