We can sorting 10 coin for 20 weighting, don't it? We can know which of the two coins is heavier for one weighting. Any two coin have different weight.
2026-03-26 12:58:30.1774529910
How to sort ten coin for twenty single weightings?
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1
I interpret the question to be whether we can determine the order of the weights of $10$ coins using $20$ weighings.
The answer is no, since there are $10!=3628800$ different orders and only $2^{20}=1048576$ different results of the $20$ weighings. You need at least $22$ weighings (and probably quite a bit more).