I thought about throwing you in the deep end on this one, but even just the last part of the puzzle is challenging enough, so I’ll lead you up to that point:

We saw from the Prime Balance puzzle (POTW #380) that if we want a number of distinct prime number blocks equally spaced around a circle to have a centre of gravity at the centre of the circle, the least number of blocks is six. This is because if you want distinct rational weights, the number of weights cannot be a prime or a prime power. If we add in a further stipulation that the number of blocks must be odd, the lowest number that is odd but neither a prime nor a prime power is 15.

Using the fact that 15 is 3 x 5 we can set up a system of equations as we did before:

A = V

B = W+T

C = X+U

D = Y

E = Z+T

F = V+U

G = W

H = X+T

I = Y+U

J = Z

K = V+T

L = W+U

M = X

N = Y+T

O = Z+U

Whatever we choose for T to Z the system of blocks will be balanced around the centre of the circle. We still have to ensure that the numbers A to O are all prime and all different. What is the least possible total weight of the 15 blocks?