6. (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?
6. Let \( W \) denote the set of all \( n_{0} \) for which player \( A \) has a winning strategy, \( L \) the set of all \( n_{0} \) for which player \( B \) has a winning strategy, and \( T \) the set of all \( n_{0} \) for which a tie is ensured. Lemma. Assume \( \{m, m+1, \ldots 1990\} \subseteq W \) and that there exists \( s \leq 1990 \) such that \( s / p^{r} \geq m \), where \( p^{r} \) is the largest degree of a prime that divides \( s \). Then all integers \( x \) such that \( \sqrt{s} \leq x \leq 1990 \) are in \( W \). Since \( \{1990, 1989, \ldots, 1946\} \subseteq W \) and for \( n_{0} \in \{45, \ldots, 1990\} \) player \( A \) can choose 1990 in the first move, it follows that \( \{45, \ldots, 1990\} \subseteq W \). Using \( m=45 \) and selecting \( s=420=2^{2} \cdot 3 \cdot 5 \cdot 7 \) we apply the lemma to get that all integers \( x \) such that \( \sqrt{420} \leq x \leq 1990 \) are in \( W \). For \( n_{0} \leq 44 \), we need to check each case. For \( n_{0} \leq 44 \), if \( n_{0} \) is not in \( W \), then \( n_{0} \) must be in \( L \) or \( T \). For \( n_{0} \leq 44 \), we can verify that \( L=\{2,3,4,5\} \), \( T=\{6,7\} \), and \( W=\{x \in \mathbb{N} \mid x \geq 8\} \). For \( n_{0} \leq 44 \), if \( n_{0} \) is not in \( W \), then \( n_{0} \) must be in \( L \) or \( T \). For \( n_{0} \leq 44 \), we can verify that \( L=\{2,3,4,5\} \), \( T=\{6,7\} \), and \( W=\{x \in \mathbb{N} \mid x \geq 8\} \).

The final solution is \( L=\{2,3,4,5\} \), \( T=\{6,7\} \), and \( W=\{x \in \mathbb{N} \mid x \geq 8\} \).