Twenty-four possible answers against 3^3 = 27 outcome sequences leaves just enough room, and four against four splits those answers into exactly 8, 8 and 8. Six against six always tips, so it wastes the level outcome and leaves twelve answers for nine remaining sequences, which makes halving impossible rather than merely slow. The counting argument bounds outcome sequences rather than strategies, so it rules out every adaptive continuation at once, and the explicit schedule closes the positive half.
Take the centre, then mirror every move through it, and you place the last coin. The proof has three requirements and only one of them needs that opening move, which is the step a one-line answer skips. Central symmetry alone is not the condition: an annulus is centrally symmetric and the first player loses on it.
Taking one coin from each bag reads 9.9 ounces whichever bag is light, and the failure is blindness rather than imprecision. Loading i coins from bag i makes the dial an injective function of the culprit, at a cost of the tenth triangular number. A 45-coin variant is cheaper, and powers of two identify any subset of light bags from one reading.
Cut two diagonally opposite corners off a chessboard and 62 = 2 x 31 stays true, yet nothing fits. Writing the colour of a square as the sign (-1)^(i+j) turns the argument into arithmetic: every domino sums to zero, the two lost corners both carried +1, and the board left over is 30 against 32. The converse, Gomory's theorem, is the harder half and it goes the other way.