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You have twelve coins. You know that one is fake. The only thing that distinguishes the fake coin from the real coins is that its weight is imperceptibly different. You have a perfectly balanced scale. The scale only tells you which side weighs more than the other side.

What is the smallest number of times you must use the scale in order to always find the fake coin?

Use only the twelve coins themselves and no others, no other weights, no cutting coins, no pencil marks on the scale. etc.

These are modern coins, so the fake coin is not necessarily lighter.

Presume the worst case scenario, and don't hope that you will pick the right coin on the first attempt.

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marked as duplicate by Gareth McCaughan, Matt, dcfyj, Techidiot, Wu33o Feb 8 '17 at 13:43

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.

  • $\begingroup$ I agree; this is almost certainly a duplicate. I've seen this problem before in several forms, including having been used in one of Infocom's text adventures (Spellbreaker). $\endgroup$ – Jeff Zeitlin Feb 8 '17 at 13:31
  • $\begingroup$ hint given to me, hope it will help u to answer -It is much smaller than you first might think. Try to solve the problem first with one poisoned bottle out of eight total bottles of wine. $\endgroup$ – Khushbu_sipl Feb 8 '17 at 13:40