Question 6
A natural number $n$ is given. Determine all $(n-1)$-tuples of nonnegative integers $a_1, a_2, \dots, a_{n-1}$ such that
$$
\left[ \frac{m}{2^n - 1} \right] + \left[ \frac{2m + a_1}{2^n - 1} \right] + \left[ \frac{2^2m + a_2}{2^n - 1} \right] + \left[ \frac{2^3m + a_3}{2^n - 1} \right] + \dots + \left[ \frac{2^{n-1}m + a_{n-1}}{2^n - 1} \right] = m
$$
holds for all $m \in \mathbb{Z}$.
$$
\left[ \frac{m}{2^n - 1} \right] + \left[ \frac{2m + a_1}{2^n - 1} \right] + \left[ \frac{2^2m + a_2}{2^n - 1} \right] + \left[ \frac{2^3m + a_3}{2^n - 1} \right] + \dots + \left[ \frac{2^{n-1}m + a_{n-1}}{2^n - 1} \right] = m
$$
holds for all $m \in \mathbb{Z}$.