Alice and Bob take turns playing the tower of babel game, with Alice starting. In this game Alice has $m$ parcels of land. In each of Alice's turns she receives $n$ blocks and decides to distribute them between her parcels to make the towers on each tower taller (at the beginning the towers have a height of $0$ blocks. In each of Bob's turns he can choose to destroy a tower and render that parcel of land unusable for the rest of the game, for what values of $m,n,k$ does Bob win and for which does Alice win? Alice wins if at any point in time there is a tower of height $k$ on the field.
Here is what I have done up to now:
Suppose the game is different, in this game Alice tries to build the highest tower possible. What is the highest tower Alice can build if both players play optimally? denote this number by $h(n,m)$
The best strategy for bob is clearly to remove the highest tower in each of his turns. the best strategy for Alice is to first make sure all towers are of size $1$, then of size $2$ and so on. Why? The tower will achieve its highest point when it is the last tower standing. The height of that tower will be $(m-1)n$ minus the number of blocks on the destroyed towers, by using the aforementioned strategy Alice can minimize that sum.
So now all we have to do is calculate $t(n,m)$, we already know $t(n,m)=(m-1)n-d$ where $d$ is the number of blocks on the destroyed towers.
Starting at the end, consider when $m = 1$. In that case, Alice puts all $n$ blocks on that one parcel and wins if $k \leq n$.
If $m = 2$, then Alice will put blocks on the two parcels, Bob will remove one parcel, and then we'll be at the $m = 1$ state again. If the two parcels are not equal, Bob will remove the largest, so Alice's strategy will be to distribute as evenly as possible. If $n$ is even, then she will put $n/2$ on each, and Alice will win if $k \leq 3n/2$.
This pattern continues: Alice will put $n/i$ blocks on each parcel when there are $i$ parcels remaining. If numbers do not divide evenly, she will arrange it so that the largest tower will have exactly one block more than the smallest. If $m > n$, she may have several turns at the start where she places one block on each of $n$ empty parcels.
The approximate maximum $k$ for which Alice can win is:
$$ k = \sum_{i=1}^m \frac ni $$
Round off will alter that a bit.