imc

The circumference of circle $A$ is $200\pi$, and the circumference of circle $B$ with radius $r$ is $2r\pi$. Since circle $B$ makes a complete revolution and ends up on the same point, the circumference of $A$ must be a multiple of the circumference of $B$, therefore the quotient must be an integer. Thus, $\frac{200\pi }{2\pi \cdot r}=\frac{100}{r}$. Therefore $r$ must then be a factor of $100$, excluding $100$ because the problem says that $r<100$. $100=2^2\cdot 5^2$. Therefore $100$ has $(2+1)\cdot (2+1)$ factors*. But you need to subtract $1$ from $9$, in order to exclude $100$. Therefore the answer is $\boxed{8}$.