2003 AMC 10A Problem 8

P1-10~2 minPrint

Divisibility & Factors Basic Probability Set up the equation/formula and compute; no special trick needed

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

Show solution without answering

Key insight

60 has 12 factors, and every whole number from 1 to 6 divides 60, so 6 of the 12 factors are less than 7.

Solution

A probability here is (factors less than $7$ ) divided by (all factors), so list the factors of $60$ carefully.

List them in pairs that multiply to $60$ , starting from $1$ and going up:

small	$1$	$2$	$3$	$4$	$5$	$6$
partner	$60$	$30$	$20$	$15$	$12$	$10$

After $6$ the next number to try is $7$ , which does not divide $60$ , and then $8$ , which does not either. The next divisor is $10$ , already listed as a partner, so the list is complete.
2. That gives $6$ pairs, so $60$ has $12$ factors.
3. The factors less than $7$ are exactly the top row: $1, 2, 3, 4, 5, 6$ . That is $6$ factors.
4. The probability is
$\frac{6}{12} = \frac{1}{2}.$

The answer is $\boxed{\textbf{(E)}\ \frac{1}{2}}$ .

Why this works

Factors come in pairs that multiply to the number, one small and one large. For $60$ every pair has its small member at most $6$ and its large member at least $10$ , so the cutoff $7$ splits the factors exactly in half. When a probability question asks about factors, listing them in pairs keeps you from missing any.

Alternative approach

Count the factors with the prime factorization. Since $60 = 2^2 \times 3 \times 5$ , a factor uses $2$ zero, one or two times ( $3$ choices), $3$ zero or one time ( $2$ choices), and $5$ zero or one time ( $2$ choices). That is $3 \times 2 \times 2 = 12$ factors. Then note that $1, 2, 3, 4, 5, 6$ all divide $60$ , giving $\frac{6}{12} = \frac{1}{2}$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Factor: $60 = 2^2 \cdot 3 \cdot 5$ . The number of positive divisors is $(2+1)(1+1)(1+1) = 12$ .

Which divisors are less than $7$ ? Check $1, 2, 3, 4, 5, 6$ : each divides $60$ ( $60 = 6 \cdot 10 = 5 \cdot 12 = 4 \cdot 15$ , and so on). That is $6$ favorable divisors.

The probability is $\dfrac{6}{12} = \dfrac{1}{2}$ .

The answer is $\boxed{\textbf{(E)}\ \frac{1}{2}}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques