1990
Problem - 2236
Given that $9^{4000}$ has $3817$ digits and has a leftmost digit $9$ (base $10$). How many of the number $9^0, 9^1, 9^2, \cdots, 9^{4000}$ have leftmost digit $9$.
Answer
184
When a number is multiplied by $9$, it will gain a new digit unless the new number starts with $9$. Because $9^{4000}$ has $3816$ digits more than $9^1$, there must be $4000-3816=\boxed{184}$ numbers starting with $9$.