2018 AMC 10B Problem 3

P1-10~2 minPrint

Basic Counting Organized listing / direct enumeration

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?

Show solution without answering

Key insight

The value only depends on which two digits get multiplied together. There are just three ways to split 1, 2, 3, 4 into two pairs.

Solution

There are $24$ ways to fill the blanks, but many of them give the same value. First find out what actually matters.

Swapping the two numbers inside a product does not change it: $3 \times 4 = 4 \times 3$ .
Swapping the two products does not change the sum either.
So the value depends only on which two digits are multiplied together. The other two digits then form the second product.
Look at where the digit $1$ goes. Its partner can be $2$ , $3$ or $4$ , and then the other pair is fixed. That gives three splits:

Pairs	Value
$1 \times 2$ and $3 \times 4$	$2 + 12 = 14$
$1 \times 3$ and $2 \times 4$	$3 + 8 = 11$
$1 \times 4$ and $2 \times 3$	$4 + 6 = 10$

The three values $14, 11, 10$ are all different, so there are $3$ different values.

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

Why this works

Before listing every case, ask which changes leave the answer the same. Here the $24$ fillings shrink to only $3$ that really differ. Then list those few cases and check that no two of them happen to give the same value.

Concise solution

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

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 $\boxed{\textbf{(B)}\ 3}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques
Introduction to Counting & Probability, Chapter 4: Committees and Combinations