2009 AMC 10A Problems/Problem 19

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Problem

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 cirle $B$ 's trip. How many possible values can $r$ have?

$\mathrm{(A)}\ 4\\qquad\mathrm{(B)}\ 8\\qquad\mathrm{(C)}\ 9\\qquad\mathrm{(D)}\ 50\\qquad\mathrm{(E)}\ 90\\qquad$

Solution

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 perfect factor of the circumference of B, therefore the quotient must be an integer.

$So\qquad\frac{200\pi}{2\pi*r} = \frac{100}{r}$

R must then be a factor of 100, excluding 100 (because then circle B would be the same size as circle A). $100\: =\: 2^2\; *\; 5^2$ . Therefore 100 has $(2+1)\; *\; (2+1)\;$ 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 $a^x\: *\: b^y\: *\: c^z\;...$ and so on, is $(x+1)(y+1)(z+1)...$ .

2009 AMC 10A (Problems • Resources)
Preceded by Problem 18	Followed by Problem 20
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2009 AMC 10A Problems/Problem 19

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