You start with 10 white chips on the table. In each move you must choose exactly two pieces and turn them over: white goes to black and black goes to white. Is it ever possible to get into a situation with exactly one black piece?
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Ten tokens and the black impossible
Here the trap is not in a long count, but in a reading that is too automatic. The problem seems to advance in movements; it actually forces you to discover which feature of the board never changes.
Hints
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- Look for an invariant: something that doesn't change with a move.
- Each move changes exactly two pieces, so the parity of the number of black pieces is preserved.
- You start with 0 black; reaching 1 black would break that parity.
Solution
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Answer: No, it is impossible. Each move changes the number of black pieces in $-2$, $0$ or $+2$. Therefore, the parity (even/odd) of the number of black pieces never changes. - You start with 0 black (even).
- You want to end with 1 quarter note (odd). That contradicts the parity invariant. Reusable idea: When an operation alters quantities 2 at a time, check invariant modulo 2 before attempting long sequences.
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