Difference between revisions of "2003 AMC 12A Problems/Problem 8"
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+ | {{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #8]] and [[2003 AMC 10A Problems|2003 AMC 10A #8]]}} | ||
== 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>? | ||
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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 1== | + | ==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} | + | So the answer is <math>\boxed{\mathrm{(E)}\ \frac{1}{2}}</math>. |
− | == Solution 2== | + | === 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} | + | 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} = \boxed{\mathrm{(E)}\ \frac{1}{2}}</math> |
== See Also == | == See Also == | ||
+ | {{AMC10 box|year=2003|ab=A|num-b=7|num-a=9}} | ||
{{AMC12 box|year=2003|ab=A|num-b=7|num-a=9}} | {{AMC12 box|year=2003|ab=A|num-b=7|num-a=9}} | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] |
Revision as of 16:54, 31 July 2011
- The following problem is from both the 2003 AMC 12A #8 and 2003 AMC 10A #8, so both problems redirect to this page.
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
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 |