Difference between revisions of "2003 AMC 12A Problems/Problem 8"
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== Problem == | == Problem == | ||
− | What is the probability that a randomly drawn positive factor of <math>60</math> is less than <math>7</math> | + | What is the probability that a randomly drawn positive factor of <math>60</math> is less than <math>7</math>? |
<math> \mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{4}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2} </math> | <math> \mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{4}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2} </math> | ||
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== See Also == | == See Also == | ||
− | + | {{AMC12 box|year=2003|ab=A|num-b=7|num-a=9}} | |
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[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] |
Revision as of 21:35, 27 June 2009
Contents
Problem
What is the probability that a randomly drawn positive factor of is less than
?
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
See Also
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 |