2009 AMC 10A Problem 19

Below is the professionally curated solution for Problem 19 of the 2009 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10A solutions, or check the answer key.

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Concepts:circumference divisibility factor counting

Difficulty rating: 1630

19.

Circle $A$ has radius $100.$ Circle $B$ has an integer radius $r \lt 100$ 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?

$4$

$8$

$9$

$50$

$90$

Solution:

The circumferences are $200\pi$ and $2\pi r,$ so the initial point of tangency returns after $\dfrac{200\pi}{2\pi r} = \dfrac{100}{r}$ rolls.

For this to be an integer greater than $1,$ $r$ must be a divisor of $100$ less than $100:$ namely $1, 2, 4, 5, 10, 20, 25,$ and $50.$ That is $8$ values.

Thus, the correct answer is B.

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