8. Let the three-digit number $n=\overline{a b c}$, if the lengths $a, b, c$ can form an isosceles (including equilateral) triangle, then the number of such three-digit numbers is $\qquad$.


8. 165 Detailed Explanation: For $a, b, c$ to form the sides of a triangle, they must all be non-zero. Thus, $a, b, c \in\{1,2, \cdots, 9\}$.
(1) If they form an equilateral triangle, let the number of such three-digit numbers be $n_{1}$. Since all three digits in the three-digit number are the same, $n_{1} = C_{8}^{1} = 9$. (2) If they form an isosceles (non-equilateral) triangle, let the number of such three-digit numbers be $n_{2}$. Since the three-digit number contains only 2 different digits, denoted as $a$ and $b$, note that the positions of the two equal sides and the base can be swapped, so the number of digit pairs $(a, b)$ is $2 C_{9}^{2}$. However, when the larger number is the base, let $a > b$, it must satisfy $b < a < 2b$. In this case, the digit pairs that cannot form a triangle are:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline$a$ & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\
\hline$b$ & 4,3 & 4,3 & 3,2 & 3,2 & 1,2 & 1,2 & 1 & 1 & \\
& 2,1 & 2,1 & 1 & 1 & & & & & \\
\hline
\end{tabular}

There are a total of 20 such cases. At the same time, for each digit pair $(a, b)$, the two digits can be placed in three positions in $C_{3}^{2}$ ways. Therefore, $n_{2} = C_{3}^{2} \left(2 C_{9}^{2} - 20\right) = 6 \left(C_{9}^{2} - 10\right) = 156$. In summary, $n = n_{1} + n_{2} = 165$.