The sum of the numbers on the faces is $2+3+4+...+9=44$.

Each number contributes towards the sum on exactly 3 vertices, so the sume of all the vertices is $3\times 44 =132$. This is shared equally over $6$ vertices so the sum of each vertex must be $132 \div 6=22$.

The sums at each vertex must be equal, so in particular $G+H+9+3=F+G+H+5$ which gives $F=7$. Then the vertex sum $F+G+J+9=22$ and, since $F=7$, we get $G+J=6$.

*This problem is taken from the UKMT Mathematical Challenges.**View the archive of all weekly problems grouped by curriculum topic*

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