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, $$\dfrac{200 \pi}{2 \pi \cdot r}= \dfrac{100}{r}$$. Therefore $$r$$ must then be a factor of $$100$$, excluding $$100$$ (because then circle $$B$$ would be the same size as circle $$A$$). $$100 =2^{2}\cdot5^{2}$$. 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 $$8$$. 
圆$$A$$的半径为$$100$$.圆$$B$$的半径为整数$$r\textless100$$，当它绕着圆$$A$$的圆周滚动一次时，仍然与圆$$A$$保持内切．这两个圆在$$B$$的行程的起点和终点有相同的切点.$$r$$可能有多少个值？
1. $$4$$ 
2. $$8$$ 
3. $$9$$ 
4. $$50$$ 
5. $$90$$ 