carlosbr wrote: Two players, A and B, play the following game: they retire coins of a pile which contains initially 2006 coins. The players play removing alternatingly, in each move, from 1 to 7 coins, each player keeps the coins that retires. If a player wishes he can pass(he doesn't retire any coin), but to do that he must pay 7 coins from the ones he retired from the pile in past moves. These 7 coins are taken to a separated box and don't interfere in the game any more. The winner is the one who retires the last coin, and A starts the game. Determine which player can win for sure, it doesn't matter how the other one plays. Show the winning strategy and explain why it works. If they won't may pass at that way: $A$ starts with taking $6$ coins and after $B$ he does $8-i$ or if $B$ and he can win by take the last after $251$ moves. At the other way: $A$ takes always $7$ coins until they reach a number, after at least $143$ moves each, he has to bring the number on $8$, or B did, but $A$ can pass at least once more, (142*14<1998), (143*14>1998), so B has to take sth and then A do $8-i$ and win. If B did earlier less then $8$, A win very simply.

Let $n(X)$ be the total of coins of player $X$. Because every number is a lose number or a win number, the better strategy to $A$ is to maximize $n(A)$. Indeed, if $n(A)=7q+r$ and $n(B)= 7l+r$ and $q>l$, $A$ will win because he can pass in a lose number and win in a win point. Thus, the better strategy to $A$ is always choose $7$. Because $B$ is not interesting in lose (=0), by symmetry he has to choose always 7. Then, after several time, the pile will have $18$ coins and $n(A)=n(B)$. Thus, $A$ can choose $7$ again, and $11$ is a lose number, because if $B$ chooses $t>3$, $A$ can easily win and if $B$ choose $x$, $A$ can choose $3-x$ putting $B$ in number $8$, and because in this time $n(A)>n(B)$, $B$ will have to choose a number $v$ and of course $8-v$ less equal to $7$. Thus $A$ always wins.