Starting from an integer $n_{0}>1$, two players, $A$ and $B$, alternately name integers $n_{1}, n_{2}, n_{3}, \ldots$, adhering to the following rules.

If $n_{2 k}$ has already been named, $A$ chooses an integer $n_{2 k+1}$ at will, such that

$$
n_{2 k} \leqq n_{2 k+1} \leqq n_{2 k}^{2}
$$

is satisfied. If $n_{2 k+1}$ has already been named, $B$ chooses an integer $n_{2 k+2}$ at will, such that

$$
\frac{n_{2 k+1}}{n_{2 k+2}}
$$

is at least a prime power. Player $A$ wins if they name 1990. Player $B$ wins if they name 1. For which values of $n_{0}$ does it hold:

a) $A$ has a winning strategy,
b) $B$ has a winning strategy,

c) neither player can force a win?
We can immediately see that $B$ must always decrease the last number spoken, while $A$ must increase it (or at least not decrease it). It is also clear that $B$ can only win if $A$ speaks a prime power before him. Since $A$ can always choose a number from an interval of the type $\left[n_{2 k}, n_{2 k}^{2}\right]$, it seems that $B$ has very few chances to win, assuming that $A$ plays with some reasonable strategy.

Therefore, it is sensible to first deal with the a) question. If $n_{0} \leq 1990 < n_{0}^{2}$, then $A$ trivially wins with the number $n_{1}=1990$, so for $45 \leq n_{0} \leq 1990$, $A$ has a winning strategy.

In the case of $1990 < n_{0}$, $A$ cannot use this method, but if $A$ can say a number $n_{1}$ such that $B$ can only respond with a number $45 \leq n_{2} \leq 1990$, then $A$ wins again, since in this case $n_{2}$ takes the role of $n_{0}$. For example, $n_{1}=47 \cdot 53=2491$—$A$ can choose this number for $1990 < n_{0} \leq 2491$—, with this $n_{2}=47$ or 53.

This proves that $A$ can win if $45 \leq n_{0} \leq 2491$. Similarly, we will prove by complete induction on $k$ that $A$ has a winning strategy if $45 \leq n_{0} \leq 2^{k-1} \cdot 2491$ (for $k$ a positive integer).

For $k=1$, we have already proven the statement, so let's assume that the statement is true for some $k \geq 1$, and we need to show that it is also true for

$$
2^{k-1} \cdot 2491 < n_{0} \leq 2^{k} \cdot 2491
$$

In this case, $A$ should choose the number $n_{1}=2^{k} \cdot 2491=2^{k} \cdot 47 \cdot 53$, clearly $n_{0} \leq n_{1} < n_{0}^{2}$. After this, $B$ can only choose an integer of the form $n_{2}=n_{1} / p^{\alpha}=\left(2^{k} \cdot 47 \cdot 53\right) / p^{\alpha}$, where $p^{\alpha}$ is a prime power. For this, $47 < n_{2} \leq \frac{n_{1}}{2}=2^{k-1} \cdot 2491$, which means, according to our inductive hypothesis, that $A$ has a winning strategy, so the statement is also true for $(k+1)$.

Summarizing our results so far, we get that $A$ has a winning strategy if $45 \leq n_{0}$.

Let's try to reduce the remaining cases where $n_{0} < 45$ to this case. $A$ needs to say a number $n_{1}$ such that $B$ can only respond with a number $45 \leq n_{2}$. For this, consider the number $n_{1}=3^{2} \cdot 5 \cdot 11=495$—$A$ can choose this number for $23 \leq n_{0} \leq 44$ because then $n_{0} \leq n_{1} \leq n_{0}^{2}$, with this $n_{2}$

$$
n_{2} \geq \min \left(3^{2} \cdot 5,3^{2} \cdot 11,5 \cdot 11\right)=3^{2} \cdot 5=45
$$

which means, based on the previous results, that $A$ has a winning strategy.

The case of $n_{0} \leq 22$ remains to be examined; for $A$ it is now sufficient to say a number $n_{1}$ such that $n_{2} \geq 23$. Let $n_{1}=2 \cdot 3 \cdot 5 \cdot 7=210$, then $n_{2} \geq 2 \cdot 3 \cdot 5=30$. Since for $15 \leq n_{0} \leq 22$, $n_{0} \leq n_{1} \leq n_{0}^{2}$ also holds, in this case $A$ also has a winning strategy.

We further "step down" with the same reduction intention: for $11 \leq n_{0} \leq 14$, let $A$'s number be $n_{1}=3 \cdot 5 \cdot 7=105$; clearly $n_{0} \leq n_{1} \leq n_{0}^{2}$, and $n_{2} \geq 3 \cdot 5=15$, which means from here $A$ can win—based on the previously proven results.

Finally, let $n_{1}=2^{2} \cdot 3 \cdot 5=60$ if $8 \leq n_{0} \leq 10$; then $n_{0} \leq n_{1} \leq n_{0}^{2}$ and $n_{2} \geq \min \left(2^{2} \cdot 3,2^{2} \cdot 5,3 \cdot 5\right)=2^{2} \cdot 3=12$, which means that in this case $A$ also has a winning strategy.

This proves that $A$ has a winning strategy if $8 \leq n_{0}$.

With this method, we cannot further reduce the possible values of $n_{0}$ from which $A$ can achieve victory. Indeed, we observe that if $n_{0} < 8$, $A$ cannot say a number $n_{