6. (1992 National High School Competition Question) Let $y=f(x)$ be a function defined on $\mathbf{R}$, and satisfies the following relations: $f(10+x)=f(10-x)$, $f(20-x)=-f(20+x)$, then the function $f(x)$ is
A. An even function and a periodic function
B. An even function but not a periodic function
C. An odd function and a periodic function
D. An odd function but not a periodic function
6. From $f(20-x)=-f(20+x)$ we get $f(20-x)+f(20+x)=0 \cdots$ (1), so the graph of the function $f(x)$ is symmetric about the point (20, 0); also, from $f(10+x)=f(10-x)$ we know that the graph of the function $f(x)$ is symmetric about the line $x=10$, indicating that the function $f(x)$ is a periodic function, and one of its periods is $T=4(m-a)=40$. Furthermore, from (1) we get $f(40-x)+f(x)=0$, which means $f(-x)+f(x)=0$, so the function $f(x)$ is also an odd function, hence the answer is C.