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Difference between revisions of "2003 AMC 12A Problems/Problem 8"

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== Problem ==
 
== Problem ==
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What is the probability that a randomly drawn positive factor of <math>60</math> is less than <math>7</math>
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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 ==
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*[[2003 AMC 12A Problems]]
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{{AMC12 box|year=2003|ab=A|num-b=7|num-a=9}}
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*[[2003 AMC 12A/Problem 7|Previous Problem]]
 
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*[[2003 AMC 12A/Problem 9|Next Problem]]
 
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]

Revision as of 20:35, 27 June 2009

Problem

What is the probability that a randomly drawn positive factor of $60$ is less than $7$?

$\mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{4}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2}$

Solution 1

For a positive number $n$ which is not a perfect square, exactly half of the positive factors will be less than $\sqrt{n}$.

Since $60$ is not a perfect square, half of the positive factors of $60$ will be less than $\sqrt{60}\approx 7.746$.

Clearly, there are no positive factors of $60$ between $7$ and $\sqrt{60}$.

Therefore half of the positive factors will be less than $7$.

So the answer is $\frac{1}{2} \Rightarrow E$.

Solution 2

Testing all numbers less than $7$, numbers $1, 2, 3, 4, 5$, and $6$ divide $60$. The prime factorization of $60$ is $2^2\cdot 3 \cdot 5$. Using the formula for the number of divisors, the total number of divisors of $60$ is $(3)(2)(2) = 12$. Therefore, our desired probability is $\frac{6}{12} = \frac{1}{2} \Rightarrow E$

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
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