math problem

[kai]

Thats the method I had in mind, I thought it was 15 for the binary to find heaviest. Then there are 4 contenders for second place so that takes 3. But the third heaviest could have been eliminated by the heaviest or the second heaviest. In the worst case where the two heaviest coins met at the end you will have all the non second place coins that the heaviest eliminated (3) and all the coints the second place eliminated (3) which will take 5 moves to sort through. Therefore I think this method takes 23.

So whats your method skates?

[Silent A]

[ QUOTE ]
In the worst case where the two heaviest coins met at the end you will have all the non second place coins that the heaviest eliminated (3) and all the coints the second place eliminated (3) which will take 5 moves to sort through. Therefore I think this method takes 23.

[/ QUOTE ]

When you weigh off the four coins to find out who is second, you can eliminate one of the coins. For example:

If the four contenders are: A, B, C, and D

weighing 1 = A vs B (assume it's A)
weighing 2 = C vs D (assume it's C)
weighing 3 = winner of 1 (A) vs. winner of 2 (C) (assume it's A)

However, the coin that lost to the loser of #3 (coin D in this case) can't be the #3 coin because it's lighter than the overall winner AND at least 2 of the 2nd place contenders.

I'm pretty sure the answer is 22.

[jay_shark]

Just label the coins randomly 1,2,3,...,15,16 . Treat it like a tournament with 8 coins on one side of the bracket and 8 coins on the other . As usually the pairings are done randomly according to their number .

ie , one side is (1,2),(3,4),(5,6),(7,8) . The other side can be paired off similarly. So there are 15 possible matches in total to determine the heaviest i.e,8+4+2+1=15 .

[jay_shark]

So what we do is take the 3 opponents from the loser finalist and pair them off with each other . So we have 3 possible choices to determine the second heaviest from that side of the bracket . Now we take this coin and pair it off with the 3 losers on the winners side of the bracket . Now we have to compare the heaviest of the 6 weighings with the loser finalist and we get 22 in total . Best case scenario is that it can be done in 21 attempts but not always !

[Silent A]

[ QUOTE ]
So what we do is take the 3 opponents from the loser finalist and pair them off with each other . So we have 3 possible choices to determine the second heaviest from that side of the bracket . Now we take this coin and pair it off with the 3 losers on the winners side of the bracket . Now we have to compare the heaviest of the 6 weighings with the loser finalist and we get 22 in total . Best case scenario is that it can be done in 21 attempts but not always !

[/ QUOTE ]

There are 4 possible 2nd place coins, and up to 5 possible 3rd place coins.

[kai]

[ QUOTE ]

When you weigh off the four coins to find out who is second, you can eliminate one of the coins.

[/ QUOTE ]

I realized that after posting but didn't have time to edit my post.

This problem was among some other really difficult problems, so it just seems like the answear we came up with here is not good enough. If I find out that there is a better way I'll let you all know.