, 2021

Solution:
This problem can be visualized as a complete binary tree with $2048$ leaves, such that each node contains the sum of its two children. Let $f(p)$ be the number of ways to fill in a binary tree with $2^{p}$ leaves and the root having value $2^{p}$. We want $f(10)$.

Since all values must be a power of $2$, we can set up the recurrence $f(p) = 2 f(p-1) + f(p-1)^{2}$. This is because we have three cases: either all of the $2^{p}$ can go to the left child of the root (in which case there are $f(p-1)$ ways because even though there's $2^{p}$ in the new root, we can treat it as $2^{p-1}$ because none of the leaves will have a value of $1$), all of it can go to the right child of the root (another $f(p-1)$ ways), or it can be split evenly ($f(p-1)^{2}$ ways).

This recursion can be shown to be $f(p) = 2^{2^{p}} - 1$ by induction. Thus, our answer is $2^{1024} - 1$ which is $15$ modulo $100$.

Solution:
The simple formula derived in the previous solution hints at a cute bijection. It turns out that the entire tree is determined by the set of leaf nodes that have non-zero value. You can see this is true: start with the root node, then only split the value each time if both subtrees have a non-zero leaf. The entire process is uniquely determined. Thus, the total number of ways is $2$ to the number of leaves, minus one for the case where all of the leaves have zero value.