Each face of a cube is assigned a different integer. Then each vertex is assigned the sum of the integer values on the faces that meet at the vertex. Finally, the vertex numbers are added. What is the largest number that must divide the final sum for every possible numbering of the faces?

Guest Jul 25, 2018

#1**0 **

**Each face of a cube is assigned a different integer. Then each vertex is assigned the sum of the integer values on the faces that meet at the vertex. Finally, the vertex numbers are added. What is the largest number that must divide the final sum for every possible numbering of the faces?**

\(\begin{array}{|c|c|} \hline \text{face} & \text{assigned with integer} \\ \hline 1 & a \\ 2 & b \\ 3 & c \\ 4 & d \\ 5 & e \\ 6 & f \\ \hline \end{array}\)

\(\begin{array}{|r|r|l|} \hline \text{Vertex} & \text{faces} & \text{sum}\\ \hline 1 & a,d,e & a+d+e \\ 2 & a,b,e & a+b+e \\ 3 & b,c,e & b+c+e \\ 4 & c,d,e & c+d+e \\ 5 & a,d,f & a+d+f \\ 6 & a,b,f & a+b+f \\ 7 & b,c,f & b+c+f \\ 8 & c,d,f & c+d+f \\ \hline &\text{sum } =&(a+d+e)+(a+b+e)+(b+c+e)+(c+d+e)\\ & &+(a+d+f)+(a+b+f)+(b+c+f)+(c+d+f) \\ & = & 4a+4b+4c+4d+4e+4f \\ & = & {\color{red}4}(a+b+c+d+e+f) \\ \hline \end{array}\)

The largest number that must divide the final sum for every possible numbering of the faces is **4**

heureka Jul 25, 2018