2. The cube of an $n$-digit number is an $m$-digit number, can $n+m=$ 2001?
2. Notice that $10^{500}$ has 501 digits, and $\left(10^{500}\right)^{3}$ has 1501 digits, totaling 2002 digits.

For any positive number $a>10^{500}, a$ and $a^{3}$ have at least 2002 digits combined.

If $a<10^{500}$, it has at most 500 digits, and $a^{3}<10^{1500}$ has at most 1500 digits. Therefore, the sum of the number of digits being 2001 is impossible.