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.