Suppose you have 5 coins: 1,2,3,4,5 grams. Than you can’t use the weighing 1+2+3=6 from the solution for the six coins. Actually you can’t divide your five coins into two groups: (1,2,3) and (4,5) with one weighing. It is not clear that the sequence is non-decreasing.

]]>An interesting case is when coins’ weights form a geometric progression with the ratio at least 2. In particular, no two groups of coins have equal weights. I just received a proof from Alexey Radul that in this case you need at least n-1 weighings:

Observe that in any weighing with geometric weights, the scales are never equal, and the heaviest coin on the scale determines the result completely. Now suppose there is a sequence of weighings that purports to verify the geometric labels. Consider the coin labeled k (for k > 1). Is there a weighing in which that coin is the highest labeled coin on the scale? If not, I claim the sequence of weighings cannot distinguish between the correct labeling, and the situation where k has been switched with the next smaller coin (say k/2). Why?

Well, every weighing that includes a coin labeled higher than k will come out the same, every weighing that includes neither k nor k/2 will come out the same, and every weighing that includes k/2 but not k will also come out the same, because in those cases k/2 was supposed to dominate, and k can dominate in its stead perfectly well. Therefore, each coin labeled k (k > 1) must appear in a weighing where it is the heaviest coin, and therefore there must be at least n-1 weighings.

Q.E.D.

https://kvant.mirror1.mccme.ru/1991/09/p70.htm ]]>

Baron Munchhausen has 8 coins weighing 1, 2, …, 8 grams that look the same. Baron knows which coin is which and wants to demonstrate to his guests that he is right . He plans to conduct one weighing on the balance scale without using anything else, so that the guests will be convinced about the weight of one of the coins. Can the baron do this?

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