# Spinsum Strategy

Here is one strategy that you could use to find a solution:

Let the four numbers in the centre four cells be $$a$$, $$b$$, $$c$$ and $$d$$. Each of these numbers contributes to two line sums. Therefore the sum of the four line sums is:

$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + a + b + c + d$$

As each line sum is the same this total must be divisible by 4.

Simplifying; $$36 + a + b + c + d$$ is a multiple of 4.

As 36 is a multiple of 4 then $$a + b + c + d$$ must also be a multiple of 4.

Choose four numbers accordingly. The smallest four would be $$1 + 2 + 3 + 6$$. Arrange these in the centre cells of the Spinsum windmill so that the two smallest are not adjacent.

Now calculate the combined total of the four line sums:

$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 1 + 2 + 3 + 6 = 48$$

So each individual line sum will be a quarter of this: $$48 \div 4 = 12$$

The remaining four numbers can now be placed to achive line sums of 12. So here is one possible solution:

 4 5 1 6 3 2 7 8

All solutions