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2012 AMC 10A Problem 8

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2012 AMC 10A Problem 8AMC 10 Step by Step

Name: Date: September 13, 2026

Systems of Equations · difficulty 2 of 5 · about 2 min · position P1-10

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

A)
$4$
B)
$5$
C)
$6$
D)
$7$
E)
$8$

Answer

Key insight

Adding all three pair sums counts each number twice, so the total is 24; subtracting each pair sum from 24 recovers the missing number.

Solution

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 $\boxed{\textbf{(D)}\ 7}$ .

Why this works

A symmetric system (every variable appears the same number of times) is best attacked by adding all the equations to find the total, then peeling off one variable at a time. This "sum then subtract" trick is faster and less error-prone than substitution.

Alternative approach

The middle number is paired with the largest in the largest pair sum and with the smallest in the smallest pair sum, so it belongs to both $12$ and $19$ but not $17$ : middle $= 24 - 17 = 7$ directly.

The trap

Reporting the largest or smallest number (12 or 5), or assuming the middle pair sum 17 belongs to the middle number.

Common mistakes

Reporting the largest or smallest number (12 or 5), or assuming the middle pair sum 17 belongs to the middle number.
Subtracting equations pairwise and making sign errors instead of using the total.

Techniques

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