The 10‑Bag Coin Riddle – How I Uncovered the Fake Bag in One Simple Weigh‑In
Published on January 23 2026
When I first stumbled upon the classic “10 bags of coins” puzzle, I thought it was just another brain‑teaser to kill an idle afternoon. Little did I know that the solution would become a favorite demonstration in my workshops on logical reasoning, probability, and the power of a single, well‑planned measurement. In this post I’ll walk you through the riddle, explain why the one‑weigh‑in method works, explore a few variations, and answer the most common questions I hear from puzzlers and educators alike.
The Riddle (Plain‑English Version)
You have ten sealed bags, each supposedly containing 100 identical gold‑coloured coins. One of the bags, however, is counterfeit: every coin inside it is slightly lighter (or heavier) than the genuine ones. Using a single weighing on a balance scale, how can you determine exactly which bag is fake?
At first glance the problem seems impossible. After all, a balance scale normally tells you only relative weight – which side is heavier – not the absolute weight of a single object. Yet the trick lies in assigning a unique weight signature to each bag before the measurement.
Below is the step‑by‑step process that I use whenever I want to illustrate the elegance of this solution.
Step‑by‑Step Solution
Step Action Why It Works
1 Number the replica bags china from 1 to 10. Creates a unique identifier for each bag.
2 From bag i, take i coins. (So 1 coin from bag 1, 2 from bag 2, …, 10 from bag 10.) The total number of coins taken from each bag is distinct.
3 Place all the selected coins on the scale at once and record the total weight W. The weight will be a linear combination of the genuine and hermes constance messenger bag replica counterfeit coin weights.
4 Compute the expected weight E assuming all coins are genuine:
[ E = (1+2+…+10) \times w_\textgenuine = 55 \times w_\textgenuine ] | If no bag were fake, the scale would read E. | | 5 | Compare W to E. The difference Δ = E – W (or W – E, depending on whether the fake is lighter or heavier). | Δ equals the number of coins taken from the fake bag multiplied by the weight discrepancy per coin (Δw). | | 6 | Divide Δ by the single‑coin weight difference Δw (which you know from the problem statement, e.g., “each counterfeit coin is 1 g lighter”). The quotient tells you the bag number. | Because only one bag contributed the discrepancy, the quotient is unique. |
Concrete Example:
Suppose each genuine coin weighs 10 g, while each counterfeit coin weighs 9 g (i.e., 1 g lighter).
Expected weight: 55 × 10 g = 550 g.
After the weighing I obtain 545 g.
Δ = 550 g – 545 g = 5 g.
Δ ÷ 1 g = 5 → the fake bag is Bag 5.
That’s it – a single measurement, a tiny amount of arithmetic, guangzhou zeal replica bags reviews bags online and the answer is crystal clear.
Why a Single Weighing Is Sufficient
The secret lies in encoding information. By taking a different number of coins from each bag, the total weight becomes a weighted sum where each bag contributes a unique coefficient (its bag number). In the language of information theory, we’re mapping ten possible states (which bag is fake) onto ten distinct weight values. One measurement gives us a value that can be decoded unambiguously.
“The best puzzles are those that force you to think about how you are counting, not just what you are counting.” – Martin Gardner, Puzzle Master
Variations Worth Exploring
I love to spice up the classic version to keep my audience on their toes. Below are three popular twists, each with a brief outline of the adapted method.
Variation Change in Premise Adapted Strategy
Heavier Fake The counterfeit coins are heavier, not lighter. Same as the original; just note that Δ = W – E will be positive.
Two Fake Bags Two bags contain counterfeit coins (different weight discrepancies). Take i coins from bag i as before, but also i² coins from bag i (or use two separate weighings). The resulting weight equation has two unknowns; solving a system yields both bag numbers.
Unknown Weight Difference You are not told how much lighter/heavier the fake coin is. Use a reference of genuine coins (e.g., weigh a small sample of known good coins) to determine Δw first, then apply the original method.
Limited Coins You may take only a total of, say, best birkin bag replica 20 coins. Choose a non‑linear set (e.g., zeal replica bags reviews binary: 1, 2, 4, 8, 5) so each bag still has a unique contribution while staying under the limit.
These variations highlight how the underlying principle – designing a unique encoding – can be stretched to accommodate richer constraints.
My Personal Takeaways
Simplicity Beats Complexity – The power of the method comes from a single, elegant idea rather than a cascade of calculations.
Preparation Is Key – In a real‑world scenario (quality control, forensic analysis) you would first calibrate the scale and verify the genuine weight.
Teaching Opportunity – I routinely use this riddle to introduce linear equations, modular arithmetic, and combinatorial encoding to students of all ages.
When I first presented the puzzle to a group of high‑school seniors, the collective “Aha!” moment was worth every minute of preparation. It reinforced the notion that mathematics is not just a set of rules but a language for gucci dionysus bag replica encoding reality.
Frequently Asked Questions (FAQ)
Question Answer
Do I need a digital scale, or will a simple balance work? A balance that can tell you the exact weight (or at least the difference) is required. A two‑pan balance that only shows “left heavy/right heavy” isn’t sufficient unless you also know the exact weight of a set of genuine coins.
What if the counterfeit coins differ by a fraction of a gram? The method still works as long as the weight difference Δw is known and the scale’s precision exceeds Δw.
Can the puzzle be solved without knowing the weight of a genuine coin? Not with a single weighing. You need either the genuine weight or the exact difference Δw to decode the bag number.
Is there a way to solve it with no weighing at all? Only if you have additional information (e.g., visual defects, magnetism) external to weight. The classic version relies solely on mass.
How many weighings would be needed if you allowed any number of fake bags? In the worst case, you’d need ⌈log₂ n⌉ weighings to identify n possible fake bags, using binary search concepts.
What real‑world applications mirror this problem? Quality‑control testing (identifying a defective batch), forensic sampling (detecting adulterated material), and network diagnostics (isolating a faulty node by unique probing).
A Quick Checklist for Solving the Riddle
Label the bags 1–10.
Select i coins from bag i.
Weigh the combined set once.
Calculate the expected weight assuming all are genuine.
Find the difference Δ.
Divide Δ by the known per‑coin discrepancy.
Identify the bag number!
Closing Thoughts
The 10‑bag coin riddle is more than a clever party trick; it’s a compact illustration of how information encoding, linear algebra, and precise measurement collaborate to solve a problem that seems, at first glance, hermione granger bag replica impossible. By walking through the solution myself, I’ve learned to appreciate the subtle elegance of designing a single experiment that extracts maximal information.
If you’ve ever been stuck on a puzzling problem, ask yourself: What unique signature can I create before I measure? In many cases, the answer is simply a matter of thinking in numbers, not just objects.
I hope this deep dive has given you the tools to both solve the classic riddle and adapt its logic to new challenges. Feel free to share your own variations in the comments – I love seeing how creative minds remix this timeless puzzle.
References & Further Reading
Gardner, M. (1959). Mathematical Puzzles and louis vuitton replica bag youtube reviews 2019 Diversions. Dover Publications.
Kordemsky, A. (1981). The Moscow Puzzles. Dover Publications – Chapter on weighing puzzles.
“Information Theory in Puzzles,” Journal of Recreational Mathematics, Vol 42, 2020.
Author’s note: I’m always open to feedback. If you tried the solution and got a different result, drop a line below – together we’ll debug the methodology!