Why does A always win in this game?

Question

I have the following question with me:

"A and B start with p = 1. Then they alternately multiply p by one of the numbers 2 to 9. The winner is the one who first reaches 1000. Who wins : A or B?"

My book tells that A always wins. However, I give steps for B to win in the following way:

Step 1: A multiplies by 2 to give 2.

Step 2: B multiplies by 9 to give 18.

Step 3: A again multiplies by 9 to give 162

Step 4: B again multiplies by 9 to give 1458

Thus crossing 1000 first, B wins the game right. Is this not a possible way in which B can proceed? Is there any problem with my interpretation of my question?

Answer 1

Winning strategy for $A$:

Start by multiplying by $6$.

$B$ must then return one of $\{12,18,24,30,36,42,48,54\}$

No matter which of those $B$ returns, $A$ can win. To see this note that all $A$ has to do is to hand $B$ a number $N$ with $56≤N≤111$. If $B$ is handed such an $N$, all the possible responses lie between $112$ and $999$ and $A$ can just multiply by $9$ for the win.

It is easy for here. If $B$ returns $12$, say, then $A$ returns $60$ and wins. If $B$ returns $54$ then $A$ returns $108$ for the win, and so on. To be specific, if $B$ returns $\{12,18,24,30,36,42,48,54\}$ then $A$ returns $\{60,72,72,60,72,84,96,108\}$ respectively.

Answer 2

The OP is correct that A has a winning strategy and that there are hypothetical lines of play in which B wins.

"Who wins : A or B?" is a common, if confusing, shorthand for "Who wins in a situation where the players play perfectly?", or "Who has a winning strategy?" (for these types of games those are equivalent).