$2 \cdot 29$ Let $T=\left\{9^{k} \mid k\right.$ be an integer, $\left.0 \leqslant k \leqslant 4000\right\}$, it is known that $9^{4000}$ has 3817 digits, and its leftmost digit is 9, how many elements in $T$ have 9 as their leftmost digit?
[Solution] First, we prove a lemma: If the leftmost digit of $9^{t}$ is 9, then the leftmost digit of $9^{t-1}$ must be 1.
Indeed, if the leftmost digit of $9^{t-1}$ is $2,3,4,5,6,7,8$, it is obvious that the leftmost digit of $9^{t-1} \times 9=9^{t}$
cannot be 9.
If the leftmost digit of $9^{t-1}$ is 9, then
$$
\begin{array}{l}
9^{t-1} \leqslant \overline{999 \cdots}, \\
9^{t}=9^{t-1} \times 9 \leqslant \overline{899 \cdots}
\end{array}
$$

Thus, the leftmost digit of $9^{t}$ cannot be 9 either, so the lemma holds.
On the other hand, if the leftmost digit of $9^{t-1}$ is $2,3,4,5,6,7,8,9$, multiplying $9^{t-1}$ by 9 will result in a carry on the left, meaning that $9^{t}$ will have more digits than $9^{t-1}$.

However, when the leftmost digit of $9^{t-1}$ is 1 and the leftmost digit of $9^{t}$ is 9, the number of digits does not increase from $9^{t-1}$ to $9^{t}$. This means that when going from $9^{t-1}$ to $9^{t}$, if the leftmost digit of $9^{t}$ is 9, the number of digits does not increase.
Since $10^{4000}$ has 4001 digits and $9^{4000}$ has 3817 digits, in the continuous calculation of powers of 9, there are
$$
4001-3817=184
$$

times without a carry.
Therefore, $9^{k}$ has 184 elements whose leftmost digit is 9.