2012 AMC 10A Problem 8

P1-10~2 minPrint

Systems of Equations Set up the equation/formula and compute; no special trick needed Exploit symmetry to reduce work or pair up objects

The sums of three whole numbers taken in pairs are $12$ , $17$ , and $19$ . What is the middle number?

Show solution without answering

Key insight

Add the three pair sums. Each number is counted twice, so the three numbers total 24. Take away each pair sum from 24 to find the number left out.

Solution

1. Call the three numbers $x$ , $y$ , $z$ , with $x + y = 12$ , $x + z = 17$ and $y + z = 19$ .
2. Every number shows up in exactly two of the three sums. So adding all three sums counts every number twice:
$12 + 17 + 19 = 48 = 2(x + y + z).$
3. Half of $48$ is $24$ , so $x + y + z = 24$ .
4. Each number is the total minus the pair that leaves it out:
- $z = 24 - (x + y) = 24 - 12 = 12$ ,
- $y = 24 - (x + z) = 24 - 17 = 7$ ,
- $x = 24 - (y + z) = 24 - 19 = 5$ .
5. The numbers are $5$ , $7$ and $12$ . Check: $5 + 7 = 12$ , $5 + 12 = 17$ , $7 + 12 = 19$ . The middle number is $7$ .

The answer is $\boxed{\textbf{(D)}\ 7}$ .

Why this works

When every unknown appears the same number of times, adding all the equations gives the total in one step. After that, each unknown is the total minus a sum you already know. This is faster than substituting one equation into another, and it leaves less room for mistakes.

Alternative approach

Think about which pair leaves out the middle number. The two smaller numbers make the smallest sum, $12$ , and the two larger numbers make the largest sum, $19$ . The pair that leaves out the middle number is the smallest plus the largest, which gives the sum in between, $17$ . So the middle number is $24 - 17 = 7$ .

Concise solution

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

Let the numbers be $x, y, z$ with $x + y = 12$ , $x + z = 17$ , $y + z = 19$ .

Adding the three equations counts every number twice:
$2(x + y + z) = 12 + 17 + 19 = 48, \quad\text{so}\quad x + y + z = 24.$

Each number is the total minus the pair sum that excludes it:
$z = 24 - 12 = 12, \qquad y = 24 - 17 = 7, \qquad x = 24 - 19 = 5.$

The numbers are $5, 7, 12$ , and the middle one is $7$ .

The answer is $\boxed{\textbf{(D)}\ 7}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Intermediate Algebra, Chapter 1: Basic Techniques for Solving Equations (§1.4)
Introduction to Counting & Probability, Chapter 3: Correcting for Overcounting (§3.4)