“The greatest pleasure in life is doing something people say you cannot do.” – Walter Bagehot
When I first saw the classic “10 bags of coins, one bag has lighter (fake) coins, you get ONE weighing” brain‑teaser, I thought it was just another trick puzzle meant for math‑club flyers. But the moment I started tinkering with paper, designer knockoffs pencil, and a (imaginary) balance scale, I realized the problem is a brilliant showcase of systematic thinking and creative measurement. In this post I’ll walk you through the entire reasoning process, share the exact steps I used, and even throw in a handy table, a short FAQ, reddit legit places plus faces replica bags and a few practical tips for anyone who loves a good mental workout.
- The Puzzle Restated (In My Own Words)
You have 10 sealed bags, each containing 100 identical‑looking coins.
One of the bags is filled with counterfeit coins that are 1 gram lighter than the genuine ones.
You have a digital scale that can be used only once.
Goal: Identify the bag with the fake coins.
The trick is to decide how many coins to pull from each bag before you step onto the scale. The weight discrepancy (or lack thereof) will point directly to the culprit bag.
- My First Intuition – “What If I Use a Different Number From Each Bag?”
My mind instantly ran to the classic solution: take 1 coin from Bag 1, 2 coins from Bag 2, …, zeal replica bags reviews michael kors bags up to 10 coins from Bag 10. If all coins were genuine, the total weight would be
[ \textWeight_\textideal = (1 + 2 + 3 + \dots + 10) \times \textweight_real ]
If, say, Bag 4 contains the lighter coins, the total weight will be 4 grams lighter (because 4 coins were taken from that bag). The scale reading’s deviation from the ideal tells you exactly which bag is fake.
It’s that simple—if you know the weight of a genuine coin. In most versions the genuine coin weighs 10 g and the counterfeit 9 g, giving a 1‑gram difference per fake coin. I’ll adopt those numbers because they make the arithmetic crystal‑clear, but the logic works for any consistent weight difference.
- Building the Plan – The Table that Saves the Day
To keep everything tidy, I drafted a quick table. Table 1 shows the number of coins taken from each bag, the expected total weight if all were genuine, and how the measured weight will deviate based on which bag is fake.
Bag # Coins taken Expected weight (g) if all genuine Weight loss if this bag is fake (g)
1 1 1 × 10 = 10 1 × 1 = 1
2 2 2 × 10 = 20 2 × 1 = 2
3 3 3 × 10 = 30 3 × 1 = 3
4 4 4 × 10 = 40 4 × 1 = 4
5 5 5 × 10 = 50 5 × 1 = 5
6 6 6 × 10 = 60 6 × 1 = 6
7 7 7 × 10 = 70 7 × 1 = 7
8 8 8 × 10 = 80 8 × 1 = 8
9 9 9 × 10 = 90 9 × 1 = 9
10 10 10 × 10 = 100 10 × 1 = 10
TOTAL 55 55 × 10 = 550 g Variable (1–10 g)
Table 1 – The “take‑N‑coins‑from‑Bag‑N” strategy.
Explanation:
Coins taken is simply the bag number (Bag 1 → 1 coin, dolce and gabbana lily bag replica Bag 2 → 2 coins, …).
Expected weight assumes every coin weighs 10 g; we sum across all bags, hermes replica bags in bulk ending with 550 g.
Weight loss equals the number of coins taken from the fake bag because each counterfeit coin is 1 g lighter.
So, after placing all 55 coins on the scale, I’m looking for a reading less than 550 g. The exact shortfall (1 g, replica bags malaysia online 2 g, …, 10 g) is the identifier of the bag.
- Walking Through a Sample Run
Imagine the scale shows 545 g. That’s 5 g shy of the ideal 550 g. According to Table 1, a 5‑gram deficit points straight to Bag 5. The logic is bullet‑proof because the weight contribution from each bag is unique.
Let’s test two more scenarios for completeness:
Measured Weight (g) Deficit (g) Fake Bag
546 4 Bag 4
549 1 Bag 1
Every possible deficit (1–10 g) maps to a single ysl loulou bag replica—no ambiguity, zeal replica bags reviews no extra weighings.
- Why This Works (The Underlying Math)
At the heart of the puzzle lies a simple linear equation:
[ \textMeasured = \sum_i=1^10 (n_i \times w_\textreal) – n_k \times d ]
( n_i ) = number of coins taken from bag i (in our plan, lanvin sugar bag replica ( n_i = i )).
( w_\textreal ) = weight of a genuine coin (10 g).
( d ) = weight difference per fake coin (1 g).
( k ) = index of the bag containing fakes.
Because each ( n_i ) is distinct, subtracting ( n_k d ) creates a unique deficit that reveals k instantly.
If you’re comfortable with algebra, replica supreme bag you can solve for k directly:
[ k = \frac\textIdeal – \textMeasuredd ]
With our numbers, Ideal = 550, d = 1, so k = Ideal – Measured. Simple as that!
- A Few Friendly Tips (List Format)
Double‑Check the Coin Count – Make sure you actually pull the right number of coins from each bag. A slip (e.g., taking 5 instead of 4 from Bag 4) throws the whole method off.
Know the Real‑Coin Weight – If the problem doesn’t give the genuine weight, you can first weigh a single coin from any bag you believe to be genuine (perhaps the first one you open).
Record the “Ideal” Weight – Write down the expected total weight before you step on the scale; this prevents mental arithmetic errors.
Use a Digital Scale – A digital readout eliminates parallax errors that can creep in with analog balances.
Label Your Coins – When you dump the 55 coins onto the scale, keep a quick tally (e.g., “Bag 1:1, Bag 2:2 …”) so you can cross‑check if needed.
- Real‑World Analogies (How This Appears Outside Puzzles)
Quality‑Control in Manufacturing: A factory might test a batch of items by taking a different count from each production line, using a single mass‑measurement to locate the line with a defect.
Network Diagnostics: Sending distinct numbers of packets to each server and measuring total latency can pinpoint a slow node—conceptually similar to our weighted‑coin approach.
Medical Screening: In certain batch‑testing protocols, a single assay that aggregates variable amounts of material from each patient can reveal which sample contains an anomaly.
The underlying principle is encoding information into a single measurement by assigning unique “weights” (in our case, coin counts) to each source.
- Frequently Asked Questions (FAQ)
Question Answer
What if the fake coins are heavier, not lighter? The same approach works; the measured weight will be more than the ideal. The excess tells you the bag number.
Can I use a balance scale (two‑pan) instead of a digital one? Yes, but you’ll need to create a reference side with an equal number of genuine coins. The principle stays the same—look for the side that tilts.
What if there are two bags with fakes? With only one weighing, you can’t uniquely identify two separate bags using the simple “1‑to‑10” scheme. You’d need a more elaborate encoding (e.g., binary‑based selections) and possibly a second measurement.
Do I have to take exactly 1‑10 coins? Could I use any other pattern? Absolutely. Any set of distinct counts works, as long as each bag’s contribution to the total weight is unique. The 1‑10 arithmetic progression is just the most convenient.
What if the weight difference isn’t 1 g but something else? The formula still applies: k = (Ideal – Measured) / d. Just substitute the real difference value for d.
Is this puzzle related to the “12‑ball problem”? Yes, both are classic “one‑measurement identification” puzzles. The 12‑ball problem uses a balance scale and requires three weighings; our coin puzzle is a simpler single‑measurement variant.
- A Closing Thought
I’ve always believed that puzzles are more than idle pastimes—they’re tiny laboratories where we test the limits of logical design. The 10‑bag coin riddle illustrates how a single piece of data, when cleverly encoded, can carry enough information to solve a multi‑variable problem. It’s a reminder that sometimes the answer isn’t about doing more (more weighings, more complex equipment) but about doing smarter (arranging the inputs so the output is self‑explanatory).
So next time you find yourself with a handful of coins, a scale, and a stubborn problem, remember the simple table, the 1‑to‑10 strategy, and the delight of turning a “one‑weigh‑only” constraint into a triumphant solution.
Happy puzzling!
— [Your Name], Puzzle Enthusiast & Amateur Mathematician
References & Further Reading
Martin Gardner, Mathematical Puzzles & Diversions, (1974).
Raymond Smullyan, The Lady or the Tiger?, Chapter on weighing puzzles.
Online: “One‑Weigh‑Only Problems” – a collection of variations on the classic coin‑bag scenario.