<s>[INST] Let $n$ such that $\displaystyle{2^{n-2005}} | n!$ Let $n$ such that $\displaystyle{2^{n-2005}} | n!$ 
Prove that this number has at most
$2005$ non-zero digits when written in base $2$.
 [/INST] there are $\sum_{k\ge 1} \lfloor \frac{n}{2^k}\rfloor$ 2s in $n!$. So we must have $n - 2005 \le \sum_{k\ge 1} \lfloor \frac{n}{2^k}\rfloor$. Now write $n = \sum_{i} 2^{\ell_i}$ in base $2$ with $\ell_1 < \ldots < \ell_m$. We have
\begin{align*}
  \left  \lfloor \frac{n}{2^k}\right\rfloor &= 
      \sum_{i:\ell_i \ge k} 2^{\ell_i - k}\\
  \sum_{k\ge 1} \left  \lfloor \frac{n}{2^k}\right\rfloor 
   &= \sum_{k\ge 1} \sum_{i:\ell_i \ge k} 2^{\ell_i - k}\\
   &= \sum_{i=1}^m \sum_{k = 1}^{\ell_i} 2^{\ell_i - k}\\
   &= \sum_{i=1}^m 2^{\ell_i} \sum_{k=1}^{\ell_i}2^{-k}\\
   &= \sum_{i=1}^m 2^{\ell_i} \left(\frac{1 - 2^{-1-\ell_i}}{1-2^{-1}} - 1\right)\\
   &= \sum_{i=1}^m 2^{\ell_i} \left(1 - 2^{-\ell_i}\right)\\
   &= \sum_{i=1}^m 2^{\ell_i} - m\\
   &= n - m
\end{align*}
So $n - 2005 \le n - m$, i. e. $m \le 2005$ as to be proved.
AB,
</s>