Difference between revisions of "2003 AMC 12A Problems/Problem 8"
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<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> | ||
− | == Solution == | + | == Solution 1== |
For a positive number <math>n</math> which is not a perfect square, exactly half of the positive factors will be less than <math>\sqrt{n}</math>. | For a positive number <math>n</math> which is not a perfect square, exactly half of the positive factors will be less than <math>\sqrt{n}</math>. | ||
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Therefore half of the positive factors will be less than <math>7</math>. | Therefore half of the positive factors will be less than <math>7</math>. | ||
− | So the answer is <math>\frac{1}{2} \Rightarrow E</math>. | + | So the answer is <math>\frac{1}{2} \Rightarrow E</math>. |
+ | |||
+ | == Solution 2== | ||
+ | Testing all numbers less than <math>7</math>, numbers <math>1, 2, 3, 4, 5</math>, and <math>6</math> divide <math>60</math>. The prime factorization of <math>60</math> is <math>2^2\cdot 3 \cdot 5</math>. Using the formula for the number of divisors, the total number of divisors of <math>60</math> is <math>(3)(2)(2) = 12</math>. Therefore, our desired probability is <math>\frac{6}{12} = \frac{1}{2} \Rightarrow E</math> | ||
== See Also == | == See Also == |
Revision as of 21:49, 31 May 2008
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