Each bag contains 100 coins, and every coin in a bag weighs either 9 g, 10 g, or 11 g. We want to determine the weight of every bag using as few weighings as possible.
Take 1, 3, 9, 27, 81 coins from the five bags. If we treat 9 g = −1, 10 g = 0, and 11 g = +1 relative to 10 g, then the weight deviation becomes a unique 5-digit balanced-ternary number.
Choose a hidden weight for each bag, then perform the single weighing.
Take 1 coin from Bag A, 3 from Bag B, 9 from Bag C, 27 from Bag D, and 81 from Bag E.
These are powers of 3, so each possible deviation can be decoded uniquely.
Write the deviation from 1210 g in balanced ternary, using digits −1, 0, +1.
| Bag | Coins taken | Digit | Meaning | Contribution |
|---|
Let each bag’s deviation from 10 g be:
d₀, d₁, d₂, d₃, d₄ ∈ {−1, 0, +1}
Then the total deviation from the baseline 1210 g is
Δ = d₀·1 + d₁·3 + d₂·9 + d₃·27 + d₄·81
This is exactly a balanced-ternary representation. Since every 5-digit balanced-ternary value is unique, all 3⁵ = 243 bag assignments produce different scale readings. So one weighing determines all five bag weights.
Also, zero weighings are impossible because there are 243 possible worlds and no information is obtained. Therefore, the minimum number of weighings is 1.