Ten Boxes, One Weighing Puzzle

Problem Ten boxes labelled 1-10 each hold 100 apples weighing 1kg each, except one box whose apples all weigh 0.9kg. Using a digital scale that may be read only once, identify the box with the lighter apples.

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• The constraint that shapes everything: one reading means the answer must be encoded in a single number, so the weighing has to make each box contribute a distinguishable amount.
• The construction: take a number of apples equal to each box's label — 1 from box 1, 2 from box 2, ... 10 from box 10 — and weigh all 55 apples together.
• The arithmetic: if every apple were 1kg the total would be 1+2+...+10 = 55kg. The shortfall divided by the 0.1kg per-apple deficit gives the label directly — 54.7kg is a 0.3kg shortfall, 0.3/0.1 = 3, so box 3.
• Why it works: the label doubles as a weight, so each box maps to a unique shortfall — the same labelling/encoding trick behind positional number systems.
• Precision assumptions worth stating: the scale must resolve 0.1kg, apples are exact, and the scale can hold 55 apples — flagging assumptions is part of the answer.
• Generalizing: N boxes need 1+2+...+N apples, so the sample grows quadratically; powers of two (1, 2, 4, 8 ...) would let you identify any subset of light boxes from one reading, which is the stronger encoding.
• Recognizing the family: this is the same idea as counterfeit-coin and defective-pill puzzles — encode identity into a measurable quantity.