10.1. Find the largest integer $k$ such that for at least one natural number $n>1000$, the number $n!=1 \cdot 2 \cdot \ldots \cdot n$ is divisible by $2^{n+k+2}$.
Answer: -3. Solution. The highest power of two that divides $n!$ is the finite sum $\left[\frac{n}{2^{1}}\right]+\left[\frac{n}{2^{2}}\right]+\ldots \leq \frac{n}{2^{1}}+\frac{n}{2^{2}}+\ldots<n$, so it does not exceed $n-1$. At the same time, the equality of this power to the value $n-1$ is achieved at powers of two (for example, when $n=2^{10}=1024$). Therefore, $(k+2)_{\max }=-1, k_{\max }=-3$.