For any given $n \ge 0$ let $d(n)$ be the number of divisors of $n$. So players have 2 options each turn, replace $n$ with $d(n)$ or $n-d(n)$. Obviously $1 \le d(n) \le n$, so $n-d(n) < n$. If $d(n)=n$ (which only happens for $n=1,2$), then the player who'se turn it is can win immediatly by replacing that $n$ with 0. So for $n=1,2$ the game is won by the player who'se turn it is and for all bigger n he has to replace it with a number smaller than n. So we can easily (in theory) find out who wins when presented with a given number by first finding out who wins when presented with 3, then who wins when presented with 4, a.s.o until we reach the given number. A player where both options $d(n)$ and $n-d(n)$ are winners for the player who is presented with them will be a looser, but a player where at least one of the options $d(n)$ and $n-d(n)$ is a loosing number is itself a winner. Fortunately, we don't have to do this manually until 1036, doing until 12 proves to be enough. A player presented with 1 or 2 wins (denoted by 1W, 2W), the player presented with 3 has the 2 options to substitute the 3 with a 1 or a 2, so he looses. One can very easily continue this: 1W, 2W, 3L, 4W, 5W, 6L, 7L, 8L, 9W, 10W, 11L, 12W. Now, we have $1036=2*2*7*37$ as the prive factorization of 1036 and hence $d(1036)=12$. Anselmo (first player) has to choose between putting 12 or 1024 on the board. As we saw above, 12 would be a winning number for Bonifacio, so to have a chance Anselmo has to put the 1024 on the board. Since $1024=2^{10}$ we have $d(1024)=11$, so Bonifacio has to choose between 11 and 1013. 11 is a loosing number, so he will choose it and make Anselmo loose. So no matter what Anselmo does, he will always loose if Bonifacio plays well.