Game Dealing with Multiplication and Winning Strategy

Question

Two players play the following game. Initially X=1. The players take turns multiplying X by any whole number from 2 to 9 (inclusive). The player who first names a number greater than 1000 wins. Which, if either, can guarantee victory in this game?

Answer 1

I will approach this with us as the hero against an opponent that will play optimally (but doesn't have to).

Working backwards $\to$ what is the smallest number that we could win with if it was our turn? $1000 \div 9 = 111.\dot{1} \implies 112 \times 9 \gt 1000$

So if we have 112 or more on our turn, we will win. How do we force the enemy to give this to us? Clearly if they have 56 or more, any number in $ \{2,3...9\}$ will multiply to give us at least 112.

Working from there, how can we force them to have at least 56? OR, a better question is: What number do we need to have to force them to have 56? If we have at least 7, we can do this. $6 \times 9 =54$ is not enough. So we win if we have between 7 and 55 on our turn.

From this point, we can force them to give us between 7 and 55 if they have on their turn: 4, 5 or 6. If they have 2 or 3 they can multiply by 2 and force us to give them between 7 and 55. So if we are playing first, we would play 4, 5 or 6 and could force a win every time.

Unfortunately, if we don't have the first move, we don't win unless they play something silly like 2 or 3. Then we can force them to have between 7 and 55, and we can win. Otherwise, an optimal player will always win with the first move.

So on the first move, play 4, 5 or 6 $\to$ then multiply whatever they give you by number that gives between 55 and 111 $\to$ then whatever they multiply by doesn't matter: just multiply the number (which will be >111) by 9 and you win!