Difference between revisions of "2009 AMC 10A Problems/Problem 19"
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− | Circle <math>A</math> has radius <math>100</math>. Circle <math>B</math> has an integer radius <math>r<100</math> and remains internally tangent to circle <math>A</math> as it rolls once around the circumference of circle <math>A</math>. The two circles have the same points of tangency at the beginning and end of | + | Circle <math>A</math> has radius <math>100</math>. Circle <math>B</math> has an integer radius <math>r<100</math> and remains internally tangent to circle <math>A</math> as it rolls once around the circumference of circle <math>A</math>. The two circles have the same points of tangency at the beginning and end of circle <math>B</math>'s trip. How many possible values can <math>r</math> have? |
<math> | <math> |
Revision as of 17:53, 21 January 2017
Problem
Circle has radius . Circle has an integer radius and remains internally tangent to circle as it rolls once around the circumference of circle . The two circles have the same points of tangency at the beginning and end of circle 's trip. How many possible values can have?
Solution
The circumference of circle A is , 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 multiple of the circumference of B, therefore the quotient must be an integer.
Thus,
Therefore 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, where and are prime numbers, is .
See Also
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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