2009 AMC 10A Problem 19

Attempt Problem 19 of the 2009 AMC 10A below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

Circle AA has radius 100.100. Circle BB has an integer radius r<100r \lt 100 and remains internally tangent to circle AA as it rolls once around the circumference of circle A.A. The two circles have the same points of tangency at the beginning and end of circle BB’s trip. How many possible values can rr have?
44
88
99
5050
9090
Answer: B
Concepts:circumferencedivisibilityfactor counting
Small hint:
Divide the circumference of circle AA by the circumference of circle BB to count the rolls
Big hint:
That number is 100r,\dfrac{100}{r}, which must be an integer greater than 11
Solution:
The circumferences are 200π200\pi and 2πr,2\pi r, so the initial point of tangency returns after 200π2πr=100r\dfrac{200\pi}{2\pi r} = \dfrac{100}{r} rolls.
For this to be an integer greater than 1,1, rr must be a divisor of 100100 less than 100:100: namely 1,2,4,5,10,20,25,1, 2, 4, 5, 10, 20, 25, and 50.50. That is 88 values.
Thus, the correct answer is B.
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