AMC 10 · 2012 · #8

Grade 6 arithmetic

Archetype Pair-Sum Invariant └ Sub-archetype Universe Total Minus Known

systems-of-equations convert-to-algebra ↑ Prerequisites: linear-equations-two-var

📏 Short solution 💡 2 insights

Problem

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

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Three whole numbers are added two at a time, giving the pair sums 12, 17, and 19. Find the middle-sized of the three numbers.

Givens: There are three whole numbers; Adding them in pairs gives three sums: 12, 17, and 19; Answer choices: (A) 4, (B) 5, (C) 6, (D) 7, (E) 8

Unknowns: The middle number when the three numbers are ordered from smallest to largest

Understand

Restated: Three whole numbers are added two at a time, giving the pair sums 12, 17, and 19. Find the middle-sized of the three numbers.

Givens: There are three whole numbers; Adding them in pairs gives three sums: 12, 17, and 19; Answer choices: (A) 4, (B) 5, (C) 6, (D) 7, (E) 8

Plan

Primary tool: #4 Introduce a Variable

Secondary: #15 Organize Information in More Ways, #7 Identify Subproblems

Tool #4 (Introduce a Variable) turns the words into three clean pair-sum equations by naming the numbers $a \le b \le c$. Tool #15 (Organize Information in More Ways) is what makes it easy: order the pair sums so each sum is matched to the pair it came from. Then Tool #7 (Identify Subproblems) splits the work into two small steps — first find the total of all three numbers, then peel off the middle one — instead of solving a full three-way system.

Execute — Answer: D

#4 Introduce a Variable 6.EE.B.6 Step 1

Name the numbers, match the sums

Call the three numbers $a \le b \le c$.
Each pair sum is one of $a+b$, $a+c$, $b+c$.
The two smallest numbers make the smallest sum and the two largest make the largest sum, so $a+b=12$, $a+c=17$, and $b+c=19$.
Notice the middle number $b$ is the one missing from the sum $a+c=17$.

$$a+b=12,\quad a+c=17,\quad b+c=19$$

💡 Giving the unknown numbers names lets you write the word clues as equations you can work with.

#7 Identify Subproblems 6.EE.B.7 Step 2

Add all three pair sums

Add the three equations.
On the left every number shows up in exactly two sums, so the total is $2(a+b+c)$.
On the right, $12+17+19=48$.
That gives $2(a+b+c)=48$, so all three numbers add to $a+b+c=24$.

$$(a+b)+(a+c)+(b+c)=2(a+b+c)=12+17+19=48 \;\Rightarrow\; a+b+c=24$$

💡 Summing all the pair sums counts each number twice, so it equals double the grand total.

#7 Identify Subproblems 4.NBT.B.4 Step 3

Peel off the middle number

The middle number $b$ is missing from the pair $a+c=17$.
Subtract that pair from the grand total: $b = (a+b+c) - (a+c) = 24 - 17 = 7$.
So the middle number is 7, which is choice (D).
(The three numbers turn out to be 5, 7, and 12, and $5+7=12$, $5+12=17$, $7+12=19$ all check.)

$$b = 24 - 17 = 7 \;\Rightarrow\; \textbf{(D)}$$

💡 Taking the total of all three and removing the two that flank the middle leaves the middle number alone.

[1] #4 6.EE.B.6 Call the three numbers $a \le b \le c$. Each pair sum is one of $a+b$, $a+c$, $b

[2] #7 6.EE.B.7 Add the three equations. On the left every number shows up in exactly two sums,

[3] #7 4.NBT.B.4 The middle number $b$ is missing from the pair $a+c=17$. Subtract that pair from

Review

Reasonableness: The recovered numbers 5, 7, 12 are whole numbers and their pair sums are $5+7=12$, $5+12=17$, $7+12=19$ — exactly the three sums given, so nothing was invented. The middle value is 7, and it sits between 5 and 12 as a middle number should. A tempting trap is 5 (choice B), which is the smallest number, not the middle one; the question asks for the middle, so 7 is correct.

Alternative: Solve the system directly. From $a+b=12$ and $b+c=19$, subtract to get $c-a=7$. Combined with $a+c=17$, adding gives $2c=24$, so $c=12$, then $a=5$ and $b=12-5=7$. Same middle number, confirming (D).

CCSS standards used (min grade 6)

6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Naming the three numbers $a \le b \le c$ and writing each pair sum as an equation.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Adding the three equations to get $2(a+b+c)=48$ and solving for the total $a+b+c=24$.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Adding $12+17+19=48$ and subtracting $24-17=7$ to isolate the middle number.)

⭐ Add all the pair sums to get double the total, then subtract the pair that skips the middle number to find it.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Name the numbers, match the sums

Add all three pair sums

Peel off the middle number

Review

CCSS standards used (min grade 6)

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