We have a two player game where you start with a number $s_0$ and then you have a goal number $g$ (with $1 < g < s_0$). Both $s_0$ and $g$ are natural numbers. Each turn you choose a new number $s_n$ following these rules:
- $s_n$ must be less than the previous number $s_{n-1}$,
- $s_n$ must always be relatively prime to $s_{n-1}$,
- The numbers you choose can never be smaller than your goal number $g$.
The player who is able to write/choose the goal number wins the game.
I've already calculated an example with $s = 27$ and $g = 15$. If player 1 chooses $s_1 = 20$ in the first round, then player 2 must choose $s_2 = 19$ or $s_2 = 17$ in the second round. Either way, player 1 can choose $s_3 = 15$ in the third round, and wins the game.
Now, let's assume $g$ is an even number. Which values of $s_0$ will guarantee that the first player wins the game? I know that as long as $s_n$ is also an even number you can never choose $g$ in your turn. So the trick seems to be to make sure your opponent ends up with an even $s_n$. Is there any way to prove that for any odd number $x$ you can always find an even number $y$ that is relative prime to $x$? In that case, just set an odd number as $s_0$ and the first player will win by picking $y$ as $s_n$ every time.
If $g$ is even and $s_0$ is odd, then player $1$ has a winning strategy.
Proof : Player $1$ starts with $s_0-1$, an even number. Player $2$ must choose an odd number. Player $1$ again decreases by $1$. This repeats until $g$ is reached.
If $s_0$ is even, then player $2$ will win because he/she always can "land" on another even number.
In short : If $g$ is even, player $1$ has a winning strategy if and only if $s_0$ is odd. Moreover, the winning strategy is terribly boring : The winning player just has to chose the current number decreased by $1$.