10.85 Let $T=\left\{9^{k} \mid k\right.$ is an integer, $\left.0 \leqslant k \leqslant 4000\right\}$, given that $9^{4000}$ has 3817 digits, and its leftmost digit is 9, how many elements in $T$ have 9 as their leftmost digit?
(8th American Mathematical Invitational, 1990)
［Solution］First, we prove that if the leftmost digit of $9^{t}$ is 9, then the leftmost digit of $9^{t-1}$ must be 1.

Otherwise, if the leftmost digit of $9^{t-1}$ is $2,3,4,5,6,7,8$, then obviously the leftmost digit of $9^{t-1} \cdot 9=9^{t}$ would be $1,2,3,4,5,6,7,8$, and it could not be 9.
If the leftmost digit of $9^{t-1}$ is 9, then
$$
\begin{array}{c}
9^{t-1} \leqslant \overline{999 \cdots}, \\
9^{t}=9^{t-1} \cdot 9 \leqslant \overline{8991 \cdots} .
\end{array}
$$

The leftmost digit would also not be 9.
At the same time, we notice that if the leftmost digit of $9^{\imath-1}$ is $2,3,4,5,6,7,8$, 9, multiplying $9^{t-1}$ by 9 would result in a carry, meaning that $9^{t}$ would 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 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 computation of powers of 9, we have
$$
4001-3817=184
$$

times without a carry.
Thus, $T$ has 184 elements whose leftmost digit is 9.