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This question already has an answer here:

Setup:

A jeweler has a pile of eighty gold bracelets - all but one have the same weight. All the bracelets look the same including the one fake gold bracelet which weighs less than a true gold bracelet.

Question:

How can the jeweler find the fake gold bracelet in four trials using only a balance (big enough)?

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marked as duplicate by Oray, JMP, humn, Gamow, Rubio Aug 26 '17 at 20:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This is slightly different because the coin weighs less, rather than just a different weight. $\endgroup$ – Dr Xorile Aug 26 '17 at 16:25
  • $\begingroup$ So, an unbeknownst instance of a generalization is a duplicate? Gotcha. $\endgroup$ – user39732 Aug 26 '17 at 20:51
  • $\begingroup$ My question: How to find fake for $N=80$ in 4 weighs? Oray's question: How to find $N$ such that it only takes 5 weighs to find fake? Where is the duplicity? $\endgroup$ – user39732 Aug 26 '17 at 20:53
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  • First trial:

    weigh 27 bracelets against another 27. If one of your sets of 27 weighs less than the other, then that set contains the fake; if they both weigh the same, then the fake must be among the remaining 26 bracelets.

  • Second trial:

    from the set of 26 or 27 bracelets which you now know contains the fake, weigh 9 against another 9. If one of your sets of 9 weighs less than the other, then that set contains the fake; if they both weigh the same, then the fake must be among the remaining 8 or 9 bracelets.

  • Third trial:

    from the set of 8 or 9 bracelets which you now know contains the fake, weigh 3 against another 3. If one of your sets of 3 weighs less than the other, then that set contains the fake; if they both weigh the same, then the fake must be among the remaining 2 or 3 bracelets.

  • Fourth trial:

    if you know your fake is one of 2, just weigh them against each other to find which is lighter. If you know it's one of 3, weigh two of them against each other: if one is lighter, it's the fake, and if they both weigh the same, the third is the fake.

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Solution:

Weigh 27 against 27. If one set is lighter, that set contains the fake. Otherwise, the remaining 26 contain the fake. From the set that contains the fake, weigh 9 against 9. Again, if one set is lighter, that set contains the fake; otherwise, the remaining 9 (or 8) do. From there, weigh 3 against 3: if one set is lighter, etc, otherwise the remaining 2 do. If only 2 bracelets remain, weigh them against each other. If three remain, weigh 1 against another: if one is lighter, etc, otherwise etc.

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    $\begingroup$ ARGH, defeated by 11 seconds for nice formatting! :-P $\endgroup$ – Rand al'Thor Aug 26 '17 at 0:09