2003 AMC 12A Problems/Problem 8

Problem

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

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 $7$ .

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 $\frac{6}{12} = \frac{1}{2} \Rightarrow E$