# Divisor/multiple game

Two players $A$ and $B$ play the following game:

Start with the set $S$ of the first 25 natural numbers: $S=\{1,2,\ldots,25\}$.

Player $A$ first picks an even number $x_0$ and removes it from $S$: We have $S:=S-\{x_0\}$.

Then they take turns (starting with $B$) picking a number $x_n\in S$ which is either divisible by $x_{n-1}$ or divides $x_{n-1}$ and removing it from $S$.

The player who can not find a number in $S$ which is a multiple or is divisble by the previous number looses.

Is there a winning strategy?

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It's a finite game, and draws are impossible, so of course there is a winning strategy (for one of the players). Perhaps what you are really asking is which player has a winning strategy, and what might a winning strategy be? – Gerry Myerson Mar 30 '12 at 0:07
I believe there is a winning strategy for the second player. See my answer. – Aryabhata Mar 30 '12 at 7:37

Second player (B) wins.

Consider the following pairing:

$2,14$
$3,15$
$4,16$
$5,25$
$6,12$
$7,21$
$8,24$
$9,18$
$10,20$
$11,22$

The left out numbers are $1,13,17,19,23$.

Now whatever number player one (A) picks, the second player (B) picks the paired number from the above pairings.

Ultimately, player one (A) will be out of numbers, and will have to pick $1$, and then player two (B) picks $23$.

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