Value of $k$ satisfying this condition In a pile you have 100 stones. A partition of the pile in $k$ piles is good if:
1) the small piles have different numbers of stones;
2) for any partition of one of the small piles in 2 smaller piles, among the $k + 1$ piles you get 2 with the same number of stones (any pile has at least 1 stone).
How can I find the maximum and minimal values of $k$ for which this is possible. 
 A: It’s fairly  easy to find the maximum possible $k$. The smallest possible sum of $k$ distinct positive integers is $\sum_{i=1}^k i = \frac{k(k+1)}{2}$. For $k=14$ this is already $105$, which is too big, so the $k$ can’t possibly be more than $13$. $\sum_{i=1}^13 i = 91$, which is too small by $9$, so try increasing the size of the largest pile from $13$ to $13 + 9 = 22$, making piles of $1,2,3,\dots,11,12$ and $22$ stones. Splitting any of these piles in two produces a smaller pile of size at most $11$, matching one of the unsplit piles. Thus, the maximum possible value of $k$ is $13$.
The minimum is fairly easily seen to be at most $10$, using piles of sizes $1,3,5,\dots,17,19$: $\sum_{i=1}^{10} (2k-1) = 100$, and splitting an odd number will always produce a smaller odd number, matching one of the existing piles. The only remaining question is whether any smaller number is good.
Let $k$ be good, with piles of sizes $n_1 < n_2 < \dots < n_k$ witnessing this fact. Consider $n_i$ for some $i>1$. Suppose first that $n_i = 2m$ is even. Then we can split the pile with $n_i$ stones in $m$ different ways, from $1+(n_i-1)$ to $m+m$. The even split $m+m$ takes care of itself, but each of the remaining $m-1$ splits has to include a pile of size $n_j$ for some $j<i$. Thus, we must have $m-1 \le i-1$, $m \le i$, and $n_i = 2m \le 2i$. If $n_i = 2m+1$ is odd, there are still $m$ ways to split the $i$-th pile, from $1+(n_i-1)$ to $m+(m+1)$, but this time all of them have to include a pile matching one of the $i-1$ smaller piles, so we must have $m \le i-1$ and hence $n_i \le 2(i-1)+1 =$  $2i-1$. In both cases we have $n_i \le 2i$.
Suppose now that $k<10$: then by the result of the last paragraph we have $$\sum\limits_{i=1}^k n_i \le 2 \sum\limits_{i=1}^k i = k(k+1) \le 9\cdot 10 = 90 < 100,$$ too few stones altogether. Thus, $10$ is the minimum possible value of $k$.
