Six positive integers are placed on the faces of a cube. For each vertex of the cube we create a number by multiplying the numbers on the sides forming that vertex. If the sum of the numbers on the vertices is 1001 what is the sum of the numbers on the faces?
Let a,b,c,d,e, and f be the six positive integers. The numbers on the vertices are abd,abc,ace,ade,bdf,def,cef, and bcf. Since they sum to 1001, it follows that (c+d)(b+e)(a+f)=1001. Now 1001=(7)(11)(13) --> the sum of the numbers on the faces is 7+11+13=31.
Posted by Dennis
on 2007-01-26 11:01:37