The students at Lyneham Primary School considered odd and even vertices.

Imagine that the total at each vertex was an odd number. Then $G + H + 9 + 3$ is odd, so $G + H$ is odd, so $G$ and $H$ are an odd number and an even number.

$F + G + H + 5$ should also be odd, so $F$ must be odd.

But $F$ and $G$ or $H$ can't both be odd, because four of $F, G, H, J$ and $K$ are $2, 4, 6$ and $8,$ so at most one of them can be odd!

So the total at each vertex must be an even number, which means that $G + H + 9 + 3$ is even, so $G + H$ is even, so $G$ and $H$ are both even (as they can't both be odd).

Then $F + G + H + 5$ should also be even, so $F$ must be odd.

Since we can have at most one odd letter, all of the others must be even, and you can check that then the total at each vertex is even.

Then $9 + 3 + K + J = 5 + 3 + K + H$, which simplifies to $4 + J = H$. So $H$ and $J$ must be either $2$ and $6$ respectively or $4$ and $8$ respectively ($F$ is odd, so it must be the only letter whose number is not in the list).

Also $G + H + 9 + 3 = 5 + 3 + K + H$, which simplifies to $G + 4 = K$. So $G$ and $K$ must be either $2$ and $6$ respectively or $4$ and $8$ respectively.

That means that $G$ and $J$ are the two smaller numbers - they are either $2$ and $4$ respectively or $4$ and $2$ respectively. So $G + J = 6$.

Ben at Lyneham Primary School solved the problem like this:

Starting in the centre and bottom vertices, $G + H + 9 + 3 = J + K + 9 + 3$.

This means $G + H = J + K$.

The only way this is possible with the numbers $2, 4, 6, 8$ and a random odd number is by having $2 + 8$ and $4 + 6$. The other number is odd because if it was even then some vertices would be odd and others would be even (see above).

So since $2 + 8$ or $4 + 6 = 10$, $10 + 12$ is the vertex total, $22$.

Then moving to the right vertex, $5 + 3 + H + K$ being $22$, $H + K = 22 - (5 + 3).$ So $H + K = 14 = 8 + 6$, leaving $G$ and $J$ with $2$ and $4$, so $G + J = 6$.

*This problem is taken from the UKMT Mathematical Challenges.*