Printable

2010 AMC 10B Problem 8

2010 AMC 10B Problem 8AMC 10 Step by Step

Name: Date: September 13, 2026

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

A ticket to a school play cost $x$ dollars, where $x$ is a whole number. A group of 9th graders buys tickets costing a total of $\$48$ , and a group of 10th graders buys tickets costing a total of $\$64$ . How many values for $x$ are possible?

A)
$1$
B)
$2$
C)
$3$
D)
$4$
E)
$5$

Answer

Key insight

The price must divide both 48 and 64, so it is a divisor of gcd(48, 64) = 16, which has five divisors.

Solution

Each group spent a whole number of tickets times $x$ dollars, so $x$ divides $48$ and $x$ divides $64$ . A common divisor of two numbers is a divisor of their greatest common divisor:
$\gcd(48, 64) = \gcd(2^4\cdot 3,\ 2^6) = 2^4 = 16.$
The divisors of $16$ are $1, 2, 4, 8, 16$ , five in all, and each one is a legitimate price (for example $x = 16$ means $3$ and $4$ tickets).

The answer is $\boxed{\textbf{(E)}\ 5}$ .

Why this works

"Divides both $m$ and $n$ " is the same as "divides $\gcd(m,n)$ ," which collapses two conditions into one. Then the count of divisors of $2^4$ is $4 + 1 = 5$ . Watch for the extreme divisors $1$ and the gcd itself; the problem says whole number, so both are allowed.

The trap

Listing the divisors of 48 or of 64 only, or forgetting that 1 and 16 both count as possible prices.

Common mistakes

Listing the divisors of 48 or of 64 only, or forgetting that 1 and 16 both count as possible prices.
Discarding $x = 1$ as an unrealistic ticket price; the problem imposes no such restriction.

Techniques

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