2009 AMC 10A Problem 19

1. When a circle rolls without slipping, the length of the path it rolls along equals the length of its own edge that has touched the path.
2. Circle $B$ rolls all the way around circle $A$, a path of length $2\pi \times 100$. Circle $B$'s edge has length $2\pi r$. So circle $B$ goes around
$$\frac{2\pi \times 100}{2\pi r} = \frac{100}{r}\ \text{times}.$$
3. Try examples. If $r = 50$, circle $B$ turns exactly $2$ times, so the starting point of $B$ is touching again at the end. If $r = 30$, it turns $3\frac13$ times, so a different point of $B$ is touching at the end.
4. So the condition is that $\frac{100}{r}$ is a whole number, which means $r$ divides $100$.
5. The divisors of $100$ are $1, 2, 4, 5, 10, 20, 25, 50, 100$. That is $9$ numbers.
6. The problem says $r < 100$, so remove $100$. That leaves $8$ values.

The answer is $\boxed{\textbf{(B)}\ 8}$.

Rolling matches lengths: the distance rolled equals the part of the circle's edge used up. So "the same point touches again" means the big circle's edge is a whole number of small circle edges long, which is a divisibility condition. Then count divisors, and read the inequality $r < 100$ carefully.

As circle $B$ rolls without slipping, the arc of $A$ it has covered always equals the arc of $B$ that has been in contact. One trip around $A$ uses up arc length $2\pi \cdot 100$ on $A$, so the same total arc of $B$ has touched $A$.

The same point of $B$ is back at the contact point exactly when this arc is a whole number of $B$'s circumferences:
$$\frac{2\pi \cdot 100}{2\pi r} = \frac{100}{r} \text{ is an integer},$$
i.e. $r$ divides $100$.

$100 = 2^2 \cdot 5^2$ has $(2+1)(2+1) = 9$ positive divisors: $1, 2, 4, 5, 10, 20, 25, 50, 100$. The condition $r < 100$ removes $100$, leaving $8$ values.

The answer is $\boxed{\textbf{(B)}\ 8}$.