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.)

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."

To find the counterfeit coin among the 9 coins using a pan balance, you need to minimize the number of times you need to use the balance.

Here's the step-by-step explanation:

1. Divide the 9 coins into three groups of 3 coins each (A, B, and C).
2. Compare group A with group B using the pan balance.

a. If group A and group B balance each other, then the counterfeit coin is in group C.
b. If group A and group B don't balance each other, then the counterfeit coin is in either group A or group B.

3. Take the lighter group (either A or B) and divide it into three individual coins (D, E, and F).
4. Compare two of the individual coins (D and E) using the pan balance.

a. If D and E balance each other, then the counterfeit coin is individual coin F.
b. If D and E don't balance each other, then the lighter of the two is the counterfeit coin.

By following these steps, you can determine the counterfeit coin using a pan balance with a minimum of two weighings.