Difference between revisions of "2003 AMC 12A Problems/Problem 8"
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Solution by franzliszt | Solution by franzliszt | ||
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+ | ===Solution 4=== | ||
+ | This following solution isn't recommended, but just list out all the divisors of <math>60</math> to find <math>12</math> divisors. <math>6</math> of these are less than <math>7,</math> so the answer is <math>\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}</math>. ~Sophia866 | ||
==Video Solution== | ==Video Solution== | ||
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~savannahsolver | ~savannahsolver | ||
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+ | https://www.youtube.com/watch?v=jpMzPl7vkxE ~David | ||
== See Also == | == See Also == |
Latest revision as of 15:40, 19 August 2023
- The following problem is from both the 2003 AMC 12A #8 and 2003 AMC 10A #8, so both problems redirect to this page.
Contents
Problem
What is the probability that a randomly drawn positive factor of is less than
?
Solution
Solution 1
For a positive number which is not a perfect square, exactly half of the positive factors will be less than
.
Since is not a perfect square, half of the positive factors of
will be less than
.
Clearly, there are no positive factors of between
and
.
Therefore half of the positive factors will be less than .
So the answer is .
Solution 2
Testing all numbers less than , numbers
, and
divide
. The prime factorization of
is
. Using the formula for the number of divisors, the total number of divisors of
is
. Therefore, our desired probability is
Solution 3
This is not too bad with casework. Notice that . Hence,
has
factors, of which
are less than
. Thus, the answer is
.
Solution by franzliszt
Solution 4
This following solution isn't recommended, but just list out all the divisors of to find
divisors.
of these are less than
so the answer is
. ~Sophia866
Video Solution
~savannahsolver
https://www.youtube.com/watch?v=jpMzPl7vkxE ~David
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 |
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.