The baron knows the weights. His guests need more than his word.
Baron Munchausen
Riddle statement
The baron has eight identical-looking coins. Everyone knows that their weights are exactly 1, 2, 3, 4, 5, 6, 7 and 8 grams, one of each. Only he knows which is which.
He has a balance scale, with no extra weights, and may use it just once. His guests can see which coins he puts on each pan and which he leaves out, but they will not accept any claim about a coin’s weight without proof.
How can he make them deduce the exact weight of at least one coin?
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Solution
The coin left off the scale can be proved to weigh 6 grams.
The baron puts the 1, 2, 3, 4 and 5 g coins on one pan and the 7 and 8 g coins on the other. He leaves out the 6 g coin. Since he knows the weights, he can make this selection.
His guests see five coins balancing two. Any five distinct coins from this collection weigh at least
1 + 2 + 3 + 4 + 5 = 15 grams.
Any two weigh at most
7 + 8 = 15 grams.
For the pans to balance, both bounds must be attained. The five coins must therefore weigh 1, 2, 3, 4 and 5 g, and the other two must weigh 7 and 8 g. The only weight left is 6 grams.
The guests still cannot distinguish the individual coins on either pan. Yet the balance has fixed the weight of the coin they never weighed: equality leaves no room for a different selection.