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Difference between revisions of "2003 AMC 12A Problems/Problem 8"
(Combined Solutions 3 and 4, which are exactly the same, and improved notation, explanations, and grammar)
 
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=== Solution 1===
 
=== Solution 1===
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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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For any positive integer <math>n</math> which is not a perfect square, exactly half of its positive factors will be less than <math>\sqrt{n}</math>, since each such factor can be paired with one that is larger than <math>\sqrt{n}</math>. (By contrast, if <math>n</math> is a perfect square, one of its factors will be exactly <math>\sqrt{n}</math>, which would therefore have to be paired with itself.)
  
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Since <math>60</math> is not a perfect square, half of the positive factors of <math>60</math> will be less than <math>\sqrt{60}\approx 7.746</math>.  
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Since <math>60</math> is indeed not a perfect square, it follows that half of its positive factors are less than <math>\sqrt{60} \approx 7.746</math>. This estimate clearly shows that there are not even any integers, let alone factors of <math>60</math>, between <math>7</math> and <math>\sqrt{60}</math>. Accordingly, exactly half of the positive factors of <math>60</math> are in fact less than <math>7</math>, so the answer is precisely <math>\boxed{\mathrm{(E)}\ \frac{1}{2}}</math>.
  
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Clearly, there are no positive factors of <math>60</math> between <math>7</math> and <math>\sqrt{60}</math>.  
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=== Solution 2===
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Testing all positive integers less than <math>7</math>, we find that <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math> all divide <math>60</math>. The prime factorization of <math>60</math> is <math>2^2 \cdot 3 \cdot 5</math>, so using the standard [[Divisor function|formula for the number of divisors]], the total number of divisors of <math>60</math> is <math>3 \cdot 2 \cdot 2 = 12</math>. Therefore, the required probability is <math>\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}</math>.
  
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Therefore half of the positive factors will be less than <math>7</math>.  
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===Solution 3 (brute force)===
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Though this is not recommended for reasons of time, one can simply write out all the factors of <math>60</math>, eventually finding that <cmath>60 = 1 \cdot 60 = 2 \cdot 30 = 3 \cdot 20 = 4 \cdot 15 = 5 \cdot 12 = 6 \cdot 10.</cmath> Hence <math>60</math> has <math>12</math> factors, of which <math>6</math> are less than <math>7</math> (namely, <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math>), so the answer is <math>\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}</math>.
  
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So the answer is <math>\boxed{\mathrm{(E)}\ \frac{1}{2}}</math>.
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==Video Solution==
  
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=== Solution 2===
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https://youtu.be/C3prgokOdHc ~savannahsolver
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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>
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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 13:06, 20 June 2025

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 any positive integer $n$ which is not a perfect square, exactly half of its positive factors will be less than $\sqrt{n}$, since each such factor can be paired with one that is larger than $\sqrt{n}$. (By contrast, if $n$ is a perfect square, one of its factors will be exactly $\sqrt{n}$, which would therefore have to be paired with itself.)

Since $60$ is indeed not a perfect square, it follows that half of its positive factors are less than $\sqrt{60} \approx 7.746$. This estimate clearly shows that there are not even any integers, let alone factors of $60$, between $7$ and $\sqrt{60}$. Accordingly, exactly half of the positive factors of $60$ are in fact less than $7$, so the answer is precisely $\boxed{\mathrm{(E)}\ \frac{1}{2}}$.

Solution 2

Testing all positive integers less than $7$, we find that $1$, $2$, $3$, $4$, $5$, and $6$ all divide $60$. The prime factorization of $60$ is $2^2 \cdot 3 \cdot 5$, so using the standard formula for the number of divisors, the total number of divisors of $60$ is $3 \cdot 2 \cdot 2 = 12$. Therefore, the required probability is $\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}$.

Solution 3 (brute force)

Though this is not recommended for reasons of time, one can simply write out all the factors of $60$, eventually finding that \[60 = 1 \cdot 60 = 2 \cdot 30 = 3 \cdot 20 = 4 \cdot 15 = 5 \cdot 12 = 6 \cdot 10.\] Hence $60$ has $12$ factors, of which $6$ are less than $7$ (namely, $1$, $2$, $3$, $4$, $5$, and $6$), so the answer is $\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}$.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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