# Difference between revisions of "2009 AMC 10A Problems/Problem 19"

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<math>So\qquad\frac{200\pi}{2\pi*r} = \frac{100}{r}</math> | <math>So\qquad\frac{200\pi}{2\pi*r} = \frac{100}{r}</math> | ||

− | R must then be a factor of 100, excluding 100 (because then circle B would be the same size as circle A). <math>100\: =\: 2^2\; *\; 5^2</math>. Therefore 100 has <math>(2+1)\; *\; (2+1)\;</math> factors*. | + | R must then be a factor of 100, excluding 100 (because then circle B would be the same size as circle A). <math>100\: =\: 2^2\; *\; 5^2</math>. Therefore 100 has <math>(2+1)\; *\; (2+1)\;</math> factors*. But you need to subtract 1 from 9, in order to exclude 100. Therefore the answer is <math>\boxed{8}</math>. |

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

## Revision as of 12:09, 14 February 2009

The circumference of circle A is 200, and the circumference of circle B with radius is . 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.

R must then be a factor of 100, excluding 100 (because then circle B would be the same size as circle A). . Therefore 100 has factors*. But you need to subtract 1 from 9, in order to exclude 100. Therefore the answer is .

- The number of factors of and so on, is .