2009 AMC 10A Problems/Problem 19

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*. Subtracting 1 from 9, so as to exclude 100, the answer is $\boxed{8}$ .

The number of factors of $a^x\: *\: b^y\: *\: c^z\;...$ and so on, is $(x+1)(y+1)(z+1)...$ .

Page

Toolbox

Search

2009 AMC 10A Problems/Problem 19