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?

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 $23$.
