Problem #50 EASY

The Lighthouse Keeper's Bottles

Scenario Logic Deduction Optimization

Problem Statement

Old Fernandez has kept the lighthouse at the tip of the Ratnagiri coast for over thirty years. He is a quiet man who spends his evenings reading, listening to the radio, and drinking exactly one small glass of local cashew feni before bed. He gets a supply boat once every six months. On his last delivery, the young supply officer made a clerical error and sent Fernandez 12 bottles of feni instead of the usual six. Fernandez immediately noticed something was wrong. He lined all 12 bottles up on his shelf and stared at them carefully. Eleven of the bottles were his regular brand — he could tell by the slight cloudiness of the liquid and the weight of the bottle. But one bottle looked and felt identical to the others on the outside, yet he suspected it was a counterfeit filled with a lighter, watered-down liquid. The supply boat was already heading back to shore. Fernandez had only one tool available — a simple balance scale that sat in the corner of the lighthouse storage room, used for weighing fish catches. The scale had two pans but no weights. He needed to find the counterfeit bottle in the fewest number of weighings possible. Fernandez has 12 bottles, one of which is lighter than the rest. Using a balance scale with no weights, what is the minimum number of weighings needed to guarantee identifying the lighter bottle?

Answer & Quick Explanation

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3 weighings. Divide 12 into groups of 4, then 4 into 2+2, then test the final pair. Dividing into thirds maximises information from each weighing. General rule: n weighings can find one lighter item among up to 3^n candidates.

Detailed Editorial Solution

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A balance scale gives 3 outcomes per weighing. With 3 weighings, you can distinguish 3^3 = 27 different scenarios. Since there are only 12 candidates, 3 weighings are sufficient. The strategy mirrors the 9-ball solution but extended to 12. Step 1: Split 12 bottles into three equal groups: Group A (bottles 1-4), Group B (bottles 5-8), Group C (bottles 9-12). Step 2: Weighing 1: Place Group A on the left pan and Group B on the right pan. Step 3: If left pan rises (lighter): the counterfeit is in Group A. If right pan rises: it is in Group B. If balanced: it is in Group C. Step 4: You now have a group of 4 suspect bottles. Split this group: weigh bottle X1 vs bottle X2 (from the suspect group). Step 5: Weighing 2: If one side rises, that bottle is lighter — found in 2 weighings. If balanced, the counterfeit is X3 or X4. Step 6: Weighing 3: Weigh X3 vs X4. Whichever side rises contains the lighter counterfeit bottle. Done in exactly 3 weighings. Key Insight: The key is dividing into thirds at each stage, not halves. Halving wastes the 'balanced' outcome — it gives you no useful information. Dividing into three groups extracts maximum information from every weighing, since all three outcomes (left light, right light, balanced) each eliminate two-thirds of the possibilities.