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In 80 coins one coin is counterfeit. What is minimum number of  weighings to find out counterfeit coin?

PS: The counterfeit coin can be heavy or lighter.

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It can't be done with weighings at all. No self-respecting counterfeiter would commit the rookie mistake of getting the mass wrong. –  Henning Makholm Oct 20 '12 at 14:26
Please specify if you know whether the coin is heavy or light. That changes the process a bit. –  Ross Millikan Oct 20 '12 at 14:38
@Ross The question doesn't state that the counterfeit coin has a different weight from the others. My guess is that the real coins are made of gold and the counterfeit one, although it weights the same, is made of brass, and can be distinguished from the real coins with no weighings at all. –  MJD Oct 20 '12 at 15:36
@MJD: Good point. I can't imagine how to do it with less than zero weighings. –  Ross Millikan Oct 20 '12 at 15:38

4 Answers 4

Hint: Consider weighing one third of the coins on each side. What will this tell you?

In fact, it can be shown that the process, the definition of which you will be led to by this hint, is optimal. Try to think of why this is the case, and to classify how many weighings you need with $n$ coins.

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Hint: Consider weighing half of the coins on each side. What will this tell you ?

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Then what is it ? if you have some solution write it in the question! –  Belgi Oct 20 '12 at 14:27
Consider weighing one third of the coins on each side. What will this tell you? –  Lord_Farin Oct 20 '12 at 15:44
@Lord_Farin - put it as an answer, very nice –  Belgi Oct 20 '12 at 15:49
  1. Three piles of 26, 26 and 28. Eliminate 2 piles by 1 weighing operation.
  2. Worst case is 28. Make 3 piles of 9, 9, 10. Again eliminate 2 piles.
  3. Worst case is 10. Make 3 piles of 3, 3 and 4. Eliminate 2 piles.
  4. Worst case is 4. 3 piles of 1,1 and 2. Eliminate 2 by weighing.
  5. Worst case is 2. requires another weighing operation.

Total number of weighing operations 5. I am working on solution using 4 operations.

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You don't eliminate two piles when the coin can be either heavy or light. You do, however, know that those in one pile can only be heavy and those in the other can only be light (if the balance doesn't) or you eliminate two piles (if the balance does) –  Ross Millikan Jun 5 '14 at 21:49

break the pile into a group of 26 each, leaving 2 lots of 26 and 2 coins in spare. Then weigh the lot of 26 coins alternatively to check for the defective one . similarly break the lot of 26 into 8 each leaving 2 coins in spare.continuing in this way you would get the counterfited coin in 8 attempts

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