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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>
  
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== Solution 1==
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==Solution==
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=== 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>.  
  
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So the answer is <math>\frac{1}{2} \Rightarrow E</math>.
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So the answer is <math>\boxed{\mathrm{(E)}\ \frac{1}{2}}</math>.
  
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== Solution 2==
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=== Solution 2===
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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>
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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 $60$ 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

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 $\boxed{\mathrm{(E)}\ \frac{1}{2}}$.

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 $\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}$

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
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