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

(2009 AMC 10A Problems/Problem 19 Solution)
 
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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*. Subtracting 1 from 9, so as to exclude 100, the answer is <math>\boxed{8}</math>.
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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$\pi$, 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 $(x+1)(y+1)(z+1)...$.
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