This puzzle has been answered mnay places online. Two weighings are sufficient. This answer is copied from one site easily found:
"Divide them into three sets of three. Weight two of the sets. From that you can determine which of the three sets contains the light coin.
Then take that set and weigh two of the coins. If the scale balances take the lighter coin, otherwise take the coin you didn't weigh."
In a group of 9 coins, one is fake and weighs slightly more. What is the fewest number
of times you need to use a pan balance to find the counterfeit coins? (A pan balance is
a scale used to measure weights, where weights are placed in two pans. The heaviest
pan will go down and the other up.)
1 answer