Originally posted by wormerI think that there is one case where 4 weighings are needed - otherwise I can do this in 3.
math is science and i need help with the following problem. There are 12 identical coins and one balancing scale. One of the coins is fake and you have only 3 tries to figure out the coin. The problem is you do not know if the fake coin is heaver or lighter than the other coins.
Divide the 12 coins into 4 piles of 3 - call them A, B, C and D. Weigh pile A against pile B. If they are the same weight then C or D contains the counterfeit. Otherwise A or B does. Weigh one of the counterfeit candidate piles against a trusted pile. If they are different then you know whether the counterfeit pile is heavier or lighter and you have narrowed down the search to three coins. Choose two of the three - if the scales balance the counterfeit is the remaining one, otherwise you know which it is as you have deduced the relative weight of the counterfeit by now.
The problem comes when you have not weighed the coins in the pile of 3 containing the counterfeit in the first 2 weighings. Suppose the counterfeit is in pile D. You´ve weighed A against B and they are the same, you´ve weighed A against C and they are the same. You know the counterfeit is one of the three coins in D, but unless you are lucky and the scales balance when you do the third comparison you are stuffed as you do not know if the coin is heavier or lighter.