math problem

[skates]

I can't think of a better method off-hand than Sholar's. However, unless I misunderstood, his method takes 22 measurements, not 21. (Second place coin has 4 choices, which takes 3 measurements, not 2). Worst case scenario leaves 5 contendors as noted for third which takes 4 measurements. I have 15 + 3 + 4. I don't do CSE, but since we're assuming the weights are all distinct, I think this is optimal.