2018 NSW: Maths Puzzle 2


2nd puzzle by Dr Harry Wiggins

There is a room with 2018 coins in it. Player 1 and player 2 take turns removing 1,2,3,4,5,6,7,8, 9 or 10 coins. The player who removes the last coin loses. Who has the winning strategy? Player 1 or 2? And what is the winning strategy for this player?
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7/30/2018 5:00 PM
If the game reaches the point where there are 11 coins left, the next player loses, as whatever he plays he cannot take all the coins, and however many coins he takes, lowers the amount of coins enough, for the other player to immediately win. Thus the number of coins to ensure a win is 2007, but the same logic can be applied to reach this score, and so on. Thus the number of coins needed to ensure a win is 2018 - some multiple  of 11. The highest multiple of 11 lower than or equal to 2018 is 2013. Thus if the first player removes 5 (2018 - 2013) coins, the number of coins left is a multiple of 11 and thus he is guaranteed a win. The strategy is, if he removes n coins, you remove 11-n coins, ensuring the number of coins is always a multiple of 11 after your turn.
Sachin Reddy Northwood School.
7/31/2018 3:02 PM
Well done to Sachin for solving and explaining it beautifully.