Two At Most

Time to post the solution to the follow-up coin problem from last time. The solution isn't hard to come by, and is basically an extension of the first move of the solution from last time. Last time the first measurement was the 1,2,3 against the 6 and since they balanced, you knew for a fact that the 6 was in fact the 6 (it is the only single coin that can weigh as much as three others). This time you weigh the 1,2,3,4,5 against the 7,8. They will come out balanced, but there is no way the left side weighs less than 15 and no way the right side weighs more than 15, so they must each be exactly those weights. This leaves the 6 uniquely identified.

Actually, this question has an interesting generalization, if we have N coins with masses 1 through N and we know their masses, how many measurements of a balance scale does it take to prove the mass of one of the coins? It is easy to prove that you can always do it in three or fewer measurements as follows: N can be written as a sum of 3 triangle numbers (this is a result of the Fermat polygonal number theorem, though for triangle numbers it was initially shown by Gauss). Call these numbers A, B, and C, assume that A ≥ B ≥ C. Weigh 1,2,3,4...k against A such that they balance (we know such a k exists, A is triangular), the balancing proves that the weights on the right has a mass at least A. Next weight 1,2,3,4,...m and A against (B+A), (B+A) being the single mass of mass B+A, and m is defined by 1+2+3+4+...+m=B. When they balance it proves that B+A is at least B+A. Finally weigh the 1,2,3,...j and (B+A) against N, where j is defined by 1+2+3+4+..+j=C. When they balance it proves that N is at least N. But N is at most N, so N is identified.

Actually, two weighings is always sufficient to prove that you can find one mass, this was proven in a paper entitled Baron Munchhausen’s Sequence, which also contains the proof from the previous paragraph. The proof is rather long though, and there is little sense in reproducing it here.