2003 AMC 12A Problems/Problem 8

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

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.

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