1990 IMO 第 6 题

(FRG 2) ${ }^{\mathrm{IMO} 5}$ Two players $A$ and $B$ play a game in which they choose numbers alternately according to the following rule: At the beginning, an initial natural number $n_{0}>1$ is given. Knowing $n_{2 k}$, player $A$ may choose any $n_{2 k+1} \in \mathbb{N}$ such that $$n_{2 k} \leq n_{2 k+1} \leq n_{2 k}^{2}$$ Then player $B$ chooses a number $n_{2 k+2} \in \mathbb{N}$ such that $$\frac{n_{2 k+1}}{n_{2 k+2}}=p^{r}$$ where $p$ is a prime number and $r \in \mathbb{N}$. It is stipulated that player $A$ wins the game if he (she) succeeds in choosing the number 1990, and player $B$ wins if he (she) succeeds in choosing 1. For which natural numbers $n_{0}$ can player $A$ manage to win the game, for which $n_{0}$ can player $B$ manage to win, and for which $n_{0}$ can players $A$ and $B$ each force a tie?

(FRG 2) ${ }^{\mathrm{IMO} 5}$ 两个玩家 $A$ 和 $B$ 玩一个游戏，他们根据以下规则交替选择数字： 一开始，给出一个初始自然数 $n_{0}>1$。知道$n_{2 k}$，玩家$A$可以选择任何$n_{2 k+1} \in \mathbb{N}$，使得$$n_{2 k} \leq n_{2 k+1} \leq n_{2 k}^{2}$$然后玩家$B$选择一个数字$n_{2 k+2} \in \mathbb{N}$，使得$$\frac{n_{2 k+1}}{n_{2 k+2}}=p^{r}$$，其中$p$是素数，$r \in \mathbb{N}$。规定玩家$A$成功选择数字1990则获胜，玩家$B$成功选择1则获胜。玩家$A$可以在哪些自然数$n_{0}$中获胜，$B$可以在哪些自然数$n_{0}$中获胜，以及$A$和$B$分别可以在哪些自然数$n_{0}$中打平？