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$X$ and $Y$ are playing a game. There are $11$ coins on the table and each player must pick up at least $1$ coin, but not more than $5$. The person picking up the last coin loses. $X$ starts. How many should he pick up to start to ensure a win no matter what strategy $Y$ employs?

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closed as off-topic by ᴡᴏʀᴅs, Davide Giraudo, Hakim, heropup, Antonio Vargas Aug 7 '14 at 21:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – ᴡᴏʀᴅs, Davide Giraudo, Hakim, heropup, Antonio Vargas
If this question can be reworded to fit the rules in the help center, please edit the question.

What do you think? Have you ever encountered a similar problem? – barto Aug 7 '14 at 19:29
  • $X$ takes $4$

  • $Y$ picks some amount: $c$

  • $X$ then takes $6-c$

  • $Y$ has no option but to take the last one.

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If $X$ takes $4$ at the start, then there will be $7$ left.

$Y$ can then take any between $1-5$. Let $y$ be the number of coins $Y$ picks up. Then $X$ can pick up $6-y$ coins and there will be one coin left on the table, which $Y$ has to pick up. We know $X$ can pick up $6-y$ coins because $1 \leq y \leq 5$, therefore $1 \leq 6-y \leq 5$

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No matter what Y needs to be left with only $1$. Since the most someone can take is $5$ and the least someone can take is $1$ then in order to win, X needs to make sure he is picking when there are between $1+1=2$ and $1+5=6$ coins left. So to do this in the least amount of steps possible, X needs to start off by taking $4$ coins. Then there will be $7$ coins left. After Y's turn, there will be guaranteed between $2$ and $6$ coins left. Then X just needs to take $5$ coins and Y will be forced to take $1$ and X wins.

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You want your opponent to have exactly 1 coin on their turn, so on your turn you could have anywhere from 2 to 6 coins. In order to force your opponent to leave you with this many coins, they should have 7 coins. So if there are eleven coins you should take 4 as your first move.

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A player facing $n$ coins can force a win iff there exists $1\le k \le 5$ such that a player facing $n-k$ coins cannot escape a loss.

Clearly, $n=1$ is a lost position. Therefore $n=2, 3, 4, 5, 6$ are won positions. Therefore $n=7$ is a lost position. Therefore $n=8, 9,10,11,12$ are won positions. One readily sees that this pattern continues and a position $n$ is lost if and only if $n\equiv 1\pmod 6$. Therefore $n=11$ is a won position - and the (only) winning move consists in going to the lost position $7$.

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  • $X$ must not reach the state of $2\dots6$ coins left on the table at $Y$'s turn

  • Otherwise, $Y$ can take $1\dots5$ coins (leaving $1$ coin on the table) and win

  • So $X$ should take $4$ coins off the table, and leave $7$ coins on the table

  • Then, however many coins $Y$ takes, $X$ can take the remaining coins minus $1$

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