Circle $A$ has radius $100$. Circle $B$ has an integer radius $r<100$ and remains internally tangent to circle $A$ as it rolls once around the circumference of circle $A$. The two circles have the same points of tangency at the beginning and end of circle $B$'s trip. How many possible values can $r$ have?
$\mathrm{(A)}\ 4\ \qquad \mathrm{(B)}\ 8\ \qquad \mathrm{(C)}\ 9\ \qquad \mathrm{(D)}\ 50\ \qquad \mathrm{(E)}\ 90\ \qquad$
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 $\boxed{8}$.

*The number of factors of $a^x\: \cdot \: b^y\: \cdot \: c^z\;...$ and so on, where $a, b,$ and $c$ are prime numbers, is $(x+1)(y+1)(z+1)...$.