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2018 AMC 10B Problem 3

2018 AMC 10B Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Counting · difficulty 1 of 5 · about 2 min · position P1-10

In the expression $\left(\underline{\qquad}\times\underline{\qquad}\right)+\left(\underline{\qquad}\times\underline{\qquad}\right)$ each blank is to be filled in with one of the digits $1,2,3,$ or $4,$ with each digit being used once. How many different values can be obtained?

A)
$2$
B)
$3$
C)
$4$
D)
$6$
E)
$24$

Answer

Key insight

The value depends only on how the four digits are split into two pairs, and there are just three pairings: 14, 11 and 10.

Solution

Swapping the two factors inside a product, or swapping the two products, does not change the value. So the value is determined only by which two digits are multiplied together, i.e. by how $\{1,2,3,4\}$ is split into two pairs. The digit $1$ can be paired with $2$ , $3$ or $4$ , giving three pairings:

$\{1,2\},\{3,4\}$ : $2 + 12 = 14$
$\{1,3\},\{2,4\}$ : $3 + 8 = 11$
$\{1,4\},\{2,3\}$ : $4 + 6 = 10$

The three values are distinct, so there are $3$ possible values.

The answer is $\boxed{\textbf{(B)}\ 3}$ .

Why this works

Before enumerating, identify the symmetries that leave the quantity unchanged; here $4! = 24$ fillings collapse to $3$ genuinely different ones. Then list the few remaining cases explicitly and check for coincidences among the values.

The trap

Counting the 24 orderings of the digits (choice E) or the 6 ways to choose the first pair, instead of the 3 distinct pairings.

Common mistakes

Counting the 24 orderings of the digits (choice E) or the 6 ways to choose the first pair, instead of the 3 distinct pairings.
Listing the pairings but not checking whether two of them give the same sum (here they do not).

Techniques

Organized listing / direct enumeration