Difference between revisions of "2009 AMC 10A Problems/Problem 19"
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Thus, <math>\frac{200\pi}{2\pi \cdot r} = \frac{100}{r}</math>. | Thus, <math>\frac{200\pi}{2\pi \cdot r} = \frac{100}{r}</math>. | ||
− | Therefore <math>r</math> must then be a factor of <math>100</math>, excluding <math>100</math> because the problem says that <math>r<100</math>. <math>100\: =\: 2^2\; \cdot \; 5^2</math>. Therefore 100 has <math>(2+1)\; \cdot \; (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>. | + | Therefore <math>r</math> must then be a factor of <math>100</math>, excluding <math>100</math> because the problem says that <math>r<100</math>. <math>100\: =\: 2^2\; \cdot \; 5^2</math>. Therefore <math>100</math> has <math>(2+1)\; \cdot \; (2+1)\;</math> factors*. But you need to subtract <math>1</math> from <math>9</math>, in order to exclude <math>100</math>. Therefore the answer is <math>\boxed{8}</math>. |
*The number of factors of <math>a^x\: \cdot \: b^y\: \cdot \: c^z\;...</math> and so on, where <math>a, b,</math> and <math>c</math> are prime numbers, is <math>(x+1)(y+1)(z+1)...</math>. | *The number of factors of <math>a^x\: \cdot \: b^y\: \cdot \: c^z\;...</math> and so on, where <math>a, b,</math> and <math>c</math> are prime numbers, is <math>(x+1)(y+1)(z+1)...</math>. | ||
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== See Also == | == See Also == |
Latest revision as of 09:57, 8 June 2021
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 is
, and the circumference of circle
with radius
is
. Since circle
makes a complete revolution and ends up on the same point, the circumference of
must be a multiple of the circumference of
, therefore the quotient must be an integer.
Thus, .
Therefore must then be a factor of
, excluding
because the problem says that
.
. Therefore
has
factors*. But you need to subtract
from
, in order to exclude
. Therefore the answer is
.
*The number of factors ofand 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.