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Three children have 10 pieces numbered from 0 to 9 on both sides. They play the following game:

-The first child chooses a piece, so a number, preserves it and passes the number on a sheet

-The second one chooses a piece, preserves it and passes on the sheet the sum of the number chosen by him and the number written on the sheet of the first child

-The third child chooses a piece, preserves it and passes on the sheet the amount of the second child's written number and the number he chooses

Then the game resumes with the first child. When the pieces are finished, they are re-inserted into the box and the game is resumed from the child that he was following. The game is won by the first child who obtain 145 on sheet.

Show that the first child can not win this game.

I think that this problem has something wrong because I denote by $A, B, C$ the children. They move in this way:$ A, B,C,A,B,C,A,B,C, A$. At this step $B$ will write on the sheet $0+1+2+...+9=45$. After that we have the sequence $B, C,A,B,C,A,B,C,A,B$. At this step $A $ will write 90 on the sheet. And the last sequence $C, A, B, C, A, B, C,A,B,C$. At the last step $C$ will write 135. If $A$ chooses 0,$ B $ chooses 1, $C $ chooses 2 and $A$ chooses 7 then $A$ wins.

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    $\begingroup$ “So a figure”? Is that a typo? Preserves “it.” What does i”it” mean here. And what sheet? Try to be careful with language. $\endgroup$ – Thomas Andrews Nov 10 '18 at 20:25
  • $\begingroup$ @ThomasAndrews I changed somethin. I hope that now is clear. $\endgroup$ – Problemsolving Nov 10 '18 at 20:35
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    $\begingroup$ A better choice of word for "preserve" is "keep." $\endgroup$ – Sean Roberson Nov 10 '18 at 21:33
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If the players not perfect players, then $A$ can indeed win, just as you show.

However, I am sure that what the problem statement is missing is that the three players are perfect players. That is: they will all try to win, and they will all make the best possible move in order for them to win. Thus, with your example, $C$ in the end would not pick $2$, but pick $9$, and win.

The good news is that you already did the hard work for this puzzle: No matter how the first $10$ rounds play out, it will always be true after 30 moves, the number $135$ is written on the paper, and it is now $A$'s turn ... can you see how, if $B$ and $C$ are perfect players, $A$ will never win?

Interesting follow-up question: Suppose the highest priority of all the players is to win, but that if they know they can't win, they will try to prevent others from winning in the hopes of not losing either. Can you see how, if all players are perfect players, $B$ and $C$ cannot win either? (indeed, this is not a game I would like to play ... I have other things to do ...)

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