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Eighteen coins are arranged in a circle. In the beginning, all eighteen coins show tails. The following two moves are allowed:

  • Simultaneously flipping over four consecutive coins (so that tails become heads, and heads become tails); for instance the coins in positions 6,7,8,9 may be flipped simultaneously.
  • Simultaneously flipping over two pairs of consecutive coins that are separated by exactly one non-flipped coin; for instance the coins in positions 5,6,8,9 may be flipped simultaneously.

Question: Is it possible to reach the situation where all eighteen coins show heads?

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It is not possible.

Color the coins alternating black and white. There are 9 black coins and 9 white coins. Every possible move flips two black coins and two white coins, so there will always be an even number of each color showing heads. This makes it impossible for all 9 of each color to be heads up.

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  • $\begingroup$ I believe the question said all 18 coins were on tails, not half-and-half. $\endgroup$
    – Sam Weaver
    Nov 1, 2015 at 15:33
  • $\begingroup$ @SamWeaver At the start, there are 0 white coins and 0 black coins heads up. If all the coins are heads up, then there are 9 white coins and 9 black coins heads up, which is impossible. $\endgroup$
    – f''
    Nov 1, 2015 at 15:36
  • $\begingroup$ Oops, my bad, I missed that you colored the coins. I was under the impression you were using the colors to represent each face of the coin. $\endgroup$
    – Sam Weaver
    Nov 1, 2015 at 15:40

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