Find the Counterfeit Bag

There are 10 bags of 100 identical coins each. In every bag but one a coin weighs 10 grams. In the single counterfeit bag every coin is off by one gram in the same direction, weighing either 9 or 11 grams. Using a digital scale that reports the exact weight, can you find the counterfeit bag in just one weighing?

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#Make each bag answer differently

If you took the same number of coins from two bags, the scale could never tell those two apart, which is the symmetry to avoid. So take a different count from each, namely $i$ coins from bag $i$ for $i = 1, \dots, 10$ . The draw uses

1 + 2 + \cdots + 10 = 55 \tag{1}

coins, which would weigh $10 \cdot 55 = 550$ grams if every coin were genuine.

#Read the deviation

Only the counterfeit bag $k$ disturbs the total. You pulled exactly $k$ of its coins, each off by one gram in a single direction, so the whole haul is off by exactly $k$ grams,

W - 550 = \pm k. \tag{2}

Drawing a different count from each bag breaks the symmetry. A clean haul of 55 coins weighs 550 grams, and only the counterfeit bag moves the needle, by exactly its index in grams. So the size of the deviation names the bag and its sign says whether those coins run light at 9 or heavy at 11.

#One weighing is enough

The magnitude $\lvert W - 550 \rvert = k$ names the counterfeit bag, and the sign settles the rest, a negative shift meaning light 9 gram coins and a positive shift meaning heavy 11 gram coins. Because the counts $1$ through $10$ are distinct, every bag leaves its own signature on the scale, so a single weighing identifies the bag and its defect together.