Nine coins look exactly alike, but one of them is slightly heavier than the other eight, which all weigh the same. Using only a two-pan balance and no weights, what is the smallest number of weighings that is certain to find the heavy coin?

Split the coins into three groups of 3 and weigh one group against another. If a pan drops the heavy coin is in that group; if they balance it is in the third group. Either way the first weighing narrows nine coins down to three. Now weigh one of those three against another and the same reasoning picks out the coin. So 2 weighings always suffice. The trap is 3, which comes from splitting into halves as in a guessing game and forgetting that a balance has three outcomes, not two.