The first thing to do is to establish the minimum number of piles.

Clearly if there was 1 pile it would not be balanced around the centre of the disc. Two piles could be balanced but only if they were equal, which is not allowed. Three piles too can only be balanced if equal. Four can only be balanced if opposite pairs are equal.

Five is a little trickier. As we saw with the star balance puzzle it is in fact possible to have five different weights balanced around the centre. It turns out that it isn’t possible for them all to be rational, so clearly can’t all be primes.

So six piles seems our best bet. If you remember back to the pentadecagon balancing puzzle, we could split the weights into subsets of pentagons and triangles. With our six piles we can do something similar, but with triangles and opposite pairs. If we call our piles a to f, we need to find w,x,y,z such that each of the following is prime:

a=x

b=y+w

c=z

d=x+w

e=y

f=z+w

x, y and z each appear in the weights of opposite pairs, and w appears in an equilateral triangle of positions, so the system will balance whatever we choose for w,x,y,z.

Since x,y and z are all primes, and adding w to each gets new primes, w must be even. Trying 2, we need a trio of primes x,y,z that when each is increased by 2 we get three other primes, with no overlap.

Without any loss of generality we can say that x<y<z. If x=3, y cannot be 5, as x+w is already 5. It can’t be 7 either as 7+2 isn’t prime. We try y=11. z cannot be 13 but could be 17.

So with these values of w,x,y,z we have values of a to f of (3,13,17,5,11,19). These total 68 coins. Try as we might with other values for w,x,y,z, we would always need more than 68. And we can also eliminate the possibility of more than six piles: 7 will not give us a rational solution, for the same reasons that 5 won’t. And any more than that, the total of the first n prime numbers already exceeds 68.

So the minimum solution is 68 coins, in piles of 3, 13, 17, 5, 11, 19.