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2003 AMC 10A Problem 8

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2003 AMC 10A Problem 8AMC 10 Step by Step

Name: Date: September 13, 2026

Divisibility & Factors · difficulty 2 of 5 · about 2 min · position P1-10

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

A)
$\frac{1}{10}$
B)
$\frac{1}{6}$
C)
$\frac{1}{4}$
D)
$\frac{1}{3}$
E)
$\frac{1}{2}$

Answer

Key insight

60 = 2^2 * 3 * 5 has 12 divisors, and every integer from 1 to 6 divides 60, so 6 of the 12 qualify.

Solution

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

Why this works

The divisor-count formula gives the sample space instantly, and the favorable set is tiny enough to check by hand. There is also a structural reason the answer is exactly one half: divisors of $60$ pair up as $d \leftrightarrow 60/d$ , one member of each pair below $\sqrt{60} \approx 7.7$ and one above, and no divisor lies strictly between $7$ and $7.7$ .

Alternative approach

Write the divisor pairs $1 \cdot 60,\ 2 \cdot 30,\ 3 \cdot 20,\ 4 \cdot 15,\ 5 \cdot 12,\ 6 \cdot 10$ . Six pairs give twelve divisors, and exactly the smaller member of each pair is below $7$ .

The trap

Miscounting the divisors of 60 (for example 10 or 11) by listing them without the (2+1)(1+1)(1+1) formula.

Common mistakes

Miscounting the divisors of 60 (for example 10 or 11) by listing them without the (2+1)(1+1)(1+1) formula.
Using "less than or equal to 7" or forgetting $1$ as a divisor.

Techniques

Set up the equation/formula and compute; no special trick needed