(5) If the odd function $y=f(x)$ defined on $\mathbf{R}$ is symmetric about the line $x=1$, and when $0<x \leqslant 1$, $f(x)=\log _{3} x$, then the sum of all real roots of the equation $f(x)=-\frac{1}{3}+f(0)$ in the interval $(0,10)$ is $\qquad$
(5) 30 Hint: Given that the graph of the function $y=f(x)$ is symmetric about the line $x=1$, and $f(x)$ is an odd function, we have
$$
f(x+2)=f(-x)=-f(x),
$$

Therefore,
$$
f(x+4)=-f(x+2)=f(x),
$$
$f(x)$ is a periodic function, and 4 is one of its periods.
Since $f(x)$ is an odd function defined on $\mathbf{R}$, we know $f(0)=0$, and the equation $f(x)=-\frac{1}{3}+f(0)$ simplifies to $f(x)=-\frac{1}{3}$.

From the graph, we can see that $f(x)=-\frac{1}{3}$ has one real root in each of the intervals $(0,1)$ and $(1,2)$, and the sum of these two roots is $2$; $f(x)=-\frac{1}{3}$ has one real root in each of the intervals $(4,5)$ and $(5,6)$, and the sum of these two roots is $10$; $f(x)=-\frac{1}{3}$ has one real root in each of the intervals $(8,9)$ and $(9,10)$, and the sum of these two roots is $18$.

Therefore, the equation $f(x)=-\frac{1}{3}+f(0)$ has 6 distinct real roots in the interval $(0,10)$, and the sum of these 6 roots is 30.