The problem involves identifying a defective product among a set of identical-looking products, where the defective one is either heavier or lighter than the rest, using a balance scale with a limited number of weighings. The classic version of this puzzle uses 12 products and three weighings. Below is a step-by-step solution for this scenario.

• There are 12 products.
• One is defective and is either heavier or lighter than the others.
• A balance scale is used, which can compare two groups of products and indicate which group is heavier, lighter, or if they balance.
• The goal is to identify the defective product and determine whether it is heavier or lighter in three weighings.

Solution Strategy

The strategy involves dividing the products into groups and using the outcomes of each weighing to narrow down the possibilities. The key is to maximize the information gained from each weighing by carefully selecting which products to weigh against each other.

Step-by-Step Solution

1. First Weighing: Weigh products 1, 2, 3, 4 against products 5, 6, 7, 8.

   • Case 1: They balance (1,2,3,4 = 5,6,7,8).
     The defective product is among the remaining group (9, 10, 11, 12). Proceed to Step 2A.
   • Case 2: They do not balance.
     Without loss of generality, assume 1,2,3,4 is heavier than 5,6,7,8 (if 5,6,7,8 is heavier, swap the labels). This means the defective product is either one of 1,2,3,4 (heavier) or one of 5,6,7,8 (lighter). Proceed to Step 2B.

2. Case 2A: First weighing balanced (defective in 9–12).

   • Second Weighing: Weigh products 9, 10, 11 against known good products (e.g., 1, 2, 3).
     • Subcase 2A.1: They balance (9,10,11 = 1,2,3).
       Product 12 is defective.
       • Third Weighing: Weigh 12 against any good product (e.g., 1).
         • If 12 is heavier, it is heavy.
         • If 12 is lighter, it is light.
     • Subcase 2A.2: 9,10,11 is heavier than 1,2,3.
       One of 9, 10, or 11 is heavy.
       • Third Weighing: Weigh 9 against 10.
         • If 9 = 10, then 11 is heavy.
         • If 9 > 10, then 9 is heavy.
         • If 10 > 9, then 10 is heavy.
     • Subcase 2A.3: 9,10,11 is lighter than 1,2,3.
       One of 9, 10, or 11 is light.
       • Third Weighing: Weigh 9 against 10.
         • If 9 = 10, then 11 is light.
         • If 9 < 10, then 9 is light.
         • If 10 < 9, then 10 is light.

3. Case 2B: First weighing did not balance (1,2,3,4 heavier than 5,6,7,8).

   • Second Weighing: Weigh products 1, 2, 5 against 3, 6, 9 (where 9 is a known good product from the first group, as it was not weighed).
     • Subcase 2B.1: They balance (1,2,5 = 3,6,9).
       The defective product is either 4 (heavy) or 7 or 8 (light).
       • Third Weighing: Weigh 7 against 8.
         • If 7 = 8, then 4 is heavy.
         • If 7 < 8, then 7 is light.
         • If 8 < 7, then 8 is light.
     • Subcase 2B.2: 1,2,5 is heavier than 3,6,9.
       The defective product is either 1 (heavy), 2 (heavy), or 6 (light).
       • Third Weighing: Weigh 1 against 2.
         • If 1 = 2, then 6 is light.
         • If 1 > 2, then 1 is heavy.
         • If 2 > 1, then 2 is heavy.
     • Subcase 2B.3: 1,2,5 is lighter than 3,6,9.
       The defective product is either 3 (heavy) or 5 (light).
       • Third Weighing: Weigh 3 against 9 (known good).
         • If 3 > 9, then 3 is heavy.
         • If 3 = 9, then 5 is light.

Explanation

• This method ensures that all 12 products are tested in exactly three weighings, regardless of which product is defective and whether it is heavier or lighter.
• The strategy works by dividing the products into groups and using the outcomes to eliminate possibilities. Each weighing is designed to maximize the information gained, reducing the number of candidates significantly.
• If the first weighing does not balance, the second weighing includes a mix of products from both sides and a known good product to isolate the defect.
• The third weighing always distinguishes between the remaining candidates with a direct comparison.

This solution is optimal for 12 products with three weighings. For different numbers of products or conditions, the approach would need adjustment, but the core logic of systematic division and comparison remains the same.