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Tricky Pearls (Posted on 2002-10-08) Difficulty: 2 of 5
You have been given 10 bags of pearls. You are told that one of the bags is full of those cheap plastic kind of pearls, but the other nine are the real deal. Naturally, you cannot tell the difference just by looking. However, you know that the fake pearls weigh 9 grams each, while the real ones are one gram heavier and weigh 10.

Armed with a very presice scale, you could weigh a pearl from each bag until you find the fake one by weight, but that would take up to 10 weighings. Can you do it in less?

See The Solution Submitted by levik    
Rating: 3.7778 (9 votes)

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Some Thoughts Question of a question | Comment 4 of 15 |
Just some thoughts about these solutions posted here:

a) Do we know for certain that there are at least 10 pearls in each bag? If not, this could limit the possibilities of what we could do. What if the bags all had different numbers of pearls with, say, at least 2 in any bag (the question does state pearlS after all, so we can assume > 1 per bag), but no specified actual value....?

b) Also, regarding remembering which pearl came from where - if they all look alike, how would we do this? What if we couldn't mark or remember them in any? For the first solution, we'd just be ending up with a pile of 55 pearls that we'd be weighing (potentially 45 if we took none from bag 1, one from bag 2, etc.) - okay, so we could know which bag had the pearls in it at the end, but we wouldn't be able to tell which of the ones being weighed came out of that bag. We have effectively lost them unless we do more weighings. Is there any way we can prevent this kind of problem by doing the weighings another way?

I don't make these suggestions to point out holes in the question or answers, just possibly as inspiration for further discussions on other ways the question could be read and tackled...
  Posted by Nick Reed on 2002-10-08 11:43:54
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