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Difference between revisions of "2003 AMC 12A Problems/Problem 8"
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=== 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} = \boxed{\mathrm{(E)}\ \frac{1}{2}}</math>
 
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>
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===Solution 3===
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This is not too bad with casework. Notice that <math>1*60=2*30=3*20=4*15=5*12=6*10=60</math>. Hence, <math>60</math> has <math>12</math> factors, of which <math>6</math> are less than <math>7</math>. Thus, the answer is <math>\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}</math>.
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Solution by franzliszt
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===Solution 4===
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This following solution isn't recommended,  but just list out all the divisors of <math>60</math> to find <math>12</math> divisors.  <math>6</math> of these are less than <math>7,</math> so the answer is <math>\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}</math>. ~Sophia866
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==Video Solution==
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https://youtu.be/C3prgokOdHc
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~savannahsolver
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https://www.youtube.com/watch?v=jpMzPl7vkxE  ~David
  
 
== See Also ==
 
== See Also ==
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[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 14:40, 19 August 2023

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

Solution 3

This is not too bad with casework. Notice that $1*60=2*30=3*20=4*15=5*12=6*10=60$. Hence, $60$ has $12$ factors, of which $6$ are less than $7$. Thus, the answer is $\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}$.

Solution by franzliszt

Solution 4

This following solution isn't recommended, but just list out all the divisors of $60$ to find $12$ divisors. $6$ of these are less than $7,$ so the answer is $\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}$. ~Sophia866

Video Solution

https://youtu.be/C3prgokOdHc

~savannahsolver

https://www.youtube.com/watch?v=jpMzPl7vkxE ~David

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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