8. For a positive integer $n$, let the sum of its digits be denoted as $s(n)$, and the product of its digits be denoted as $p(n)$. If $s(n) + p(n) = n$ holds, then $n$ is called a coincidence number. The sum of all coincidence numbers is $\qquad$ .
Answer 531.
Analysis Let $n=\overline{a_{1} a_{2} \cdots a_{k}}, a_{i} \neq 0$, from $n-s(n)=p(n)$ we get
$$
a_{1}\left(10^{k-1}-1\right)+a_{2}\left(10^{k-2}-1\right)+\cdots+a_{k-1}(10-1)=a_{1} \cdot a_{2} \cdots a_{k}
$$

That is
$$
a_{1}\left(10^{k-1}-1-a_{2} \cdot a_{3} \cdots a_{k}\right)+m=0
$$

where $m=a_{2}\left(10^{k-2}-1\right)+\cdots \geq 0$, if $k \geq 3$, since $10^{k-1}-1-a_{2} \cdot a_{3} \cdots a_{k} \geq 10^{k-1}-1-9^{k-1}>0$, which contradicts (1), so $k \leq 2$, and when $k=1$, $n=s(n)$, which does not meet the condition, thus $k=2$, hence we can set $n=\overline{a_{1} a_{2}}=10 a_{1}+a_{2}$, then from $a_{1}+a_{2}+a_{1} a_{2}=10 a_{1}+a_{2}$, we get $9 a_{1}=a_{1} a_{2}$, so $a_{2}=9, a_{1} \in\{1,2, \cdots, 9\}$, that is, all coincidental numbers are $19,29, \cdots, 99$, their sum is 531.