Find the Counterfeit Coin: 9, 10, and 12-Coin Variants Explained

Find the Counterfeit Coin: 10 Coins (or N Coins), One Different, Minimum Weighings

Counterfeit coin puzzles are a classic interview category. The setup: among N otherwise-identical coins, exactly one differs in weight. Using a balance scale, find the counterfeit in as few weighings as possible. Variants include “the counterfeit is heavier,” “lighter,” or “different (could be either)” — each yields different optimal strategies. This guide covers the standard 9-coin problem (find heavier in 2 weighings), the 12-coin problem (find different in 3 weighings), and the general N-coin pattern using ternary search.

Variant 1: 9 Coins, One Heavier, Two Weighings

You have 9 coins. One is heavier. Find it in 2 weighings on a balance scale.

Solution

Divide into 3 groups of 3 coins each.

Weighing 1: Put group A (3 coins) on the left, group B (3 coins) on the right. Group C sits aside.

• If A > B: heavier coin is in A.
• If A < B: heavier coin is in B.
• If A == B: heavier coin is in C.

Weighing 2: Take the identified group of 3 coins. Put one on the left, one on the right, one aside.

• If left > right: left is the counterfeit.
• If left < right: right is the counterfeit.
• If equal: the aside coin is the counterfeit.

Total weighings: 2. The trick is recognizing that each weighing has 3 outcomes (left heavy, right heavy, balanced), so 3² = 9 possibilities can be distinguished in 2 weighings.

Variant 2: 10 Coins, One Different, Find It

10 coins; one is different (could be heavier or lighter). The unknown direction makes this harder than 9 coins with a known direction. With 10 coins and unknown direction, you can solve in 3 weighings.

The general formula

With known-direction counterfeit: maximum N solvable in k weighings is 3^k.

• k = 1 → 3 coins
• k = 2 → 9 coins
• k = 3 → 27 coins

With unknown-direction counterfeit: maximum N solvable in k weighings is (3^k – 3) / 2.

• k = 2 → 3 coins
• k = 3 → 12 coins
• k = 4 → 39 coins

So 10 coins with unknown direction is solvable in 3 weighings (since (3^3 – 3) / 2 = 12 ≥ 10).

Variant 3: 12 Coins, One Different, Three Weighings

This is the canonical “hard” variant. Strategy:

Label coins 1 through 12. Weighing 1: put {1, 2, 3, 4} on the left, {5, 6, 7, 8} on the right. {9, 10, 11, 12} aside.

Three outcomes from Weighing 1:

Case A: Left = Right (balanced)

Counterfeit is among {9, 10, 11, 12}. Direction unknown.

Weighing 2: put {9, 10, 11} on left, {1, 2, 3} (known good) on right.

• Balanced: counterfeit is 12. Weigh against any known good to determine heavier/lighter.
• Left heavier: counterfeit is in {9, 10, 11} and is heavier. Weighing 3: put 9 vs 10. Result tells you which (or 11 if balanced).
• Left lighter: counterfeit is in {9, 10, 11} and is lighter. Same Weighing 3 logic.

Case B: Left > Right

Counterfeit is heavy in left, or light in right. Specifically: counterfeit is in {1, 2, 3, 4} and heavy, OR in {5, 6, 7, 8} and light.

Weighing 2: A clever rearrangement. Put {1, 2, 5} on left, {3, 6, 9} (9 is known good from case A reasoning… wait, we’re in case B; 9 is unknown). Let’s restate: put {1, 2, 5} on left, {3, 4, 9} on right (where 9 is known good since case A excluded {9,10,11,12} as “all good or counterfeit” but we’re in B so {9,10,11,12} are known good).

Three outcomes give a 6-into-3 reduction; weighing 3 isolates the single counterfeit.

Case C: Left < Right

Symmetric to Case B; counterfeit is light in left or heavy in right.

The full 12-coin solution is a textbook example of ternary information-theoretic optimization. Working through it on paper is a strong interview signal.

The General Pattern: Ternary Search

Each weighing has three outcomes, distinguishing 3 possibilities. With k weighings, you can distinguish at most 3^k cases. For known-direction, each coin is one case → 3^k coins. For unknown-direction, each coin gives two cases (heavier or lighter) plus we need one “all good” outcome → roughly (3^k – 1) / 2 coins; with care, (3^k – 3) / 2.

The key insight: balance scales output ternary (left, right, equal), so the natural search is ternary, not binary. Don’t bisect — trisect.

Common Variations

Find the counterfeit and determine if it’s heavier or lighter

Same as the standard problem; the weighing strategy already determines direction in the unknown-direction variant.

Multiple counterfeits

“K of N coins are counterfeit.” Becomes a much harder combinatorial problem. Group testing literature applies (e.g., Chinese remainder theorem inspired strategies).

Coins of unknown weight

“Coins are different weights but you don’t know which is the standard.” Reduces to identifying the majority-weight or finding pairs that match.

Common Mistakes

• Bisecting instead of trisecting. The natural reflex is binary search, but balance scales give ternary outcomes. Wasted information per weighing.
• Forgetting the “balanced” outcome. Each weighing has three outcomes. Plan for all three branches.
• Using known-good coins inefficiently. After each weighing, some coins are known to be good. Use them to make subsequent weighings more informative.
• Not handling the unknown-direction case. “The counterfeit is heavier or lighter” requires more weighings than “the counterfeit is heavier” — the unknown direction roughly doubles the information needed.
• Confusing the strategy when coins move sides. In the 12-coin case B, some coins from the first weighing’s right side appear on the second weighing’s left side. The intent is to break the symmetry of “heavy on left vs light on right” — coins that change sides give clean information.

Frequently Asked Questions

See also: Bridge Crossing Puzzle • Fruit Jar Labeling Puzzle • Two-Eggs / 80-Floor Building