Two players start with $1$ and take turns multiplying it by $2$ or $3$ or $4$ ... or $9$. The first player to make the number $\ge 1000$ wins. Who has a winning strategy?
My attempt: obviously $112$ is a losing position since $112*9=1008$. Then it's a winning position for all numbers between $56$ to $111$. Then the first player can choose to multiply by $5$ on the first turn, then on his second turn he can always multiply a number so it's between $56$ and $111$. Therefore the first player has a winning strategy.
Is this solution correct. Is there another way of doing it? Also, in general, who has a winning strategy if the first player to make the number $\ge n$ wins?
Yes, this is a good strategy and demonstration that it works. $4$ and $6$ also work.
You might compare with the game where you add instead of multiply and are allowed to add $1,2,3,4$ at each turn and the first to reach some total, say $27$ wins. A player can always guarantee that two successive turns add to $5$, so the first player wins by taking $2$, then hitting $7,12,17,22,27$.
In your game a player can guarantee that two turns multiply by at least $18$ but one turn cannot multiply by more than $9$. Your $56$ comes from $\lceil \frac {1000}{18}\rceil$ and $111$ from $\lfloor \frac {1000}9 \rfloor$. This suggests that for target $T$ and $n$ pairs of turns before the end you should leave between $\lceil \frac T{18^n} \rceil$ and $\lfloor \frac T{18^{n-1}\cdot 9}\rfloor$ The worry is that you can't guarantee that two turns will multiply by exactly $18$. If your opponent multiplies by $7$ you have to choose between $2$ and $3$, making the pair multiply by $14$ or $21$. I believe you are OK because the range is a factor $2$. In this example if multiplying by $3$ is too much, $2$ will be enough to stay in range.