This classic logic challenge asks: Can you identify a single defective ball—either heavier or lighter—from a set of 12, using only three weighings on a balance scale?

The answer is yes, but it requires a precise and clever weighing strategy.

Puzzle Summary

  • You have 12 balls, visually identical.
  • One is defective: heavier or lighter, unknown which.
  • You have three weighings on a balance scale (which shows heavier side or balance).
  • Your goal: Identify the odd ball and whether it is heavier or lighter.

High-Level Strategy

Each weighing on a balance scale gives 3 outcomes: left heavier, right heavier, or balanced.
With 3 weighings, you have at most \( 3^3 = 27 \) unique outcome sequences.

There are 12 possible balls × 2 defect types (heavier/lighter) = 24 possibilities to distinguish.

So 27 > 24 — enough outcomes to identify the culprit in 3 weighings.

Step-by-Step Solution

First Weighing:

Weigh balls 1, 2, 3, 4 vs. 5, 6, 7, 8

Three outcomes:

1. If balanced:

The defective ball is among 9, 10, 11, 12, and balls 1–8 are known to be normal.

Second Weighing: Weigh 9, 10, 11 vs. 1, 2, 3 (known normal).

  • If balanced: Ball 12 is the odd one.
    • Third Weighing: Weigh 12 vs. 1. If 12 goes down, it’s heavy; if it goes up, it’s light.
  • If left is heavy: The odd ball is 9, 10, or 11, and it is heavy.
    • Third Weighing: Weigh 9 vs. 10. If 9 goes down, 9 is heavy. If 10 goes down, 10 is heavy. If balanced, 11 is heavy.
  • If left is light: The odd ball is 9, 10, or 11, and it is light.
    • Third Weighing: Weigh 9 vs. 10. If 9 goes up, 9 is light. If 10 goes up, 10 is light. If balanced, 11 is light.

2. If unbalanced (e.g., left is heavier):

Assume left (1, 2, 3, 4) is heavier than right (5, 6, 7, 8). This means either one of {1, 2, 3, 4} is heavy, or one of {5, 6, 7, 8} is light. Balls 9–12 are known normal.

Second Weighing: Weigh 1, 2, 3, 5 vs. 4, 9, 10, 11. (We kept 1, 2, 3 on the left, moved 4 to the right, moved 5 to the left, and filled the rest with normal balls).

  • If left is heavy again: The odd ball must be one that stayed on the same side and was suspected heavy. So it’s 1, 2, or 3 (all heavy candidates).
    • Third Weighing: Weigh 1 vs. 2. If 1 is heavy, it goes down. If 2 is heavy, it goes down. If balanced, 3 is heavy.
  • If right is heavy: The imbalance flipped. The odd ball must be one that switched sides. So it’s either 4 (switched to right, making it heavy) or 5 (switched to left, making it light).
    • Third Weighing: Weigh 4 vs. 9 (normal). If 4 goes down, 4 is heavy. If balanced, 5 is light.
  • If balanced: The odd ball must be one of the unweighed balls from the suspected light group: 6, 7, or 8 (all light candidates).
    • Third Weighing: Weigh 6 vs. 7. If 6 goes up, 6 is light. If 7 goes up, 7 is light. If balanced, 8 is light.

(If the first weighing resulted in the left being lighter, you apply the exact same logic but mirror the heavy/light expectations).

Final Answer

Yes, you can always identify the defective ball and its nature (heavier/lighter) in exactly 3 weighings.

This puzzle illustrates information theory and the power of base-3 logic: each weighing gives 3 possible outcomes. Three weighings give \( 3^3 = 27 \) possible outcomes, which is just enough to distinguish the 24 possible states (12 balls × 2 states).

Reference