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There are (10,2)=45 ways to pick the two coins out of the ten. Each use of the scale gives three possible result. Thus, with less than 4 uses, you cannot distinguish among 45 cases. Now, you need an algorithm that does this, to complete the proof, and there are such solutions among the comments... Extra note: if the fake coins didn't weigh the same, we would have 90 ways to pick the two coins, so the scale would have to be used 5 times. Once again, this is a minimum, and until you find the algorithm, you cannot say it's the actual minimum. |
