13. Let $T=\left\{9^{4} \mid k\right.$ be an integer, $0 \leqslant k$ $\leqslant 4000\}$. It is known that $9^{1000}$ has 3817 digits, and its most significant digit is 9. How many elements in $T$ have 9 as their most significant digit?
13. Notice that $9^{k}$ has one more digit than $9^{k-1}$, unless $9^{1}$ starts with the digit 9. In the latter case, by long division, $9^{k-1}$ starts with the digit 1, and $9^{k}$ has the same number of digits. Therefore, from $9^{0}$ to $9^{4000}$, there are 3816 digit increases, so there must be 184 times when the number of digits from $9^{k-1}$ to $9^{k}$ does not increase.
$\because 9^{\circ}=1$, the leading digit is not 9,
$\therefore$ when the number of digits does not increase from $9^{\mathrm{k}-1}$ to $9^{\text { }}$, $9^{\mathbf{k}}(1 \leqslant k$ $\leqslant 4000$ ) has 9 as the leading digit.
Thus, there are 184 numbers that have 9 as the leftmost digit.