44. There are $n k$ stones, which are somehow distributed into $n$ piles. It is allowed to double any pile by moving stones from other piles into it in any way. For which $k$ can such operations always (for any $n$) ensure that all remaining piles have an equal number of stones?
92.44. Answer. Only for $k$ being a power of two. Arrange $2k$ stones into two piles, one containing 1 stone and the other containing $2k-1$ stones. Suppose we can obtain two identical piles of $k$ stones each (or one pile of $2k$ stones, which is the same). What does the operation described in the condition look like? The pair $(x, y)$ transitions to the pair $(2x, y-x)$. The reverse operation looks like this: the pair $(a, b)$ transitions to the pair $(a/2, b + a/2)$. Clearly, moving in the reverse direction, we can only obtain pairs of the form $(ra + sb, (1-r)a + (1-s)b)$, where $r$ and $s$ are proper fractions with a denominator that is a power of two. On the other hand, if the pair $(k, k)$ can be transformed into the pair $(1, 2k-1)$, then $1 = k(r + s)$, which implies that $k$ is a power of two.

The second part of the proof is left to the reader.