To identify the faulty weight from 9 equal weights using a two-pan balance, we can apply a strategy similar to a "divide and conquer" method, where the weights are split into groups for comparison.

Strategy:

1. First weighing: Divide the 9 weights into three groups of 3 (Group A, Group B, and Group C). Weigh Group A against Group B.

   • If they balance, the faulty weight is in Group C.
   • If they don't balance, the faulty weight is in the heavier or lighter group.

2. Second weighing: Now, take the group identified from the first weighing (whether Group A, B, or C) and divide it into three individual weights. Weigh two of the weights against each other.

   • If they balance, the third weight is the faulty one.
   • If they don’t balance, the heavier or lighter one is the faulty weight.

Minimum Number of Weighings:

• First weighing: Reduces the problem from 9 weights to 3.
• Second weighing: Identifies the faulty weight out of 3.

Thus, 2 weighings are the minimum required to find the faulty weight.

Summary:

The least number of weighings required to identify the faulty weight out of 9 is 2.

Would you like further details or examples?

Here are some related questions:

1. How does this strategy change with a different number of weights?
2. What if the balance can only detect heavier, not lighter, objects?
3. Can we identify the faulty weight with more weighings in a different way?
4. What is the reasoning behind dividing the weights into groups of 3?
5. How does this method scale for larger numbers, like 27 or 81 weights?

Tip: When using a balance, dividing the problem into smaller chunks often simplifies the solution process.