2003 AMC 12A Problems/Problem 8

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 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 $\frac{1}{2} \Rightarrow E$ .

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

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
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2003 AMC 12A Problems/Problem 8

Contents

Problem

Solution 1

Solution 2

See Also