Problem - 2720
A positive integer is written on each face of a cube. Then for each vertex of the cube, the product of the numbers on the three faces associated with this vertex is calculated. If the sum of these eight products equals 2015, find the sum of all the numbers on the 6 faces.
Let $A$, $B$, $C$, $D$, $E$, and $F$ be the six numbers written on each face, respectively. Then we have
$$ABC+ACD+ADE+AEB+FBC+FCD+FDE+FEB=(A+F)(B+D)(C+E)=2015$$
Because $2015$ can be uniquely factorized as $2015= 5\times 13\times 31$, hence, we have $$A+F+B+D+C+E=5+13+31=\boxed{49}$$