AMC 8 · 2004 · #11

Grade 6 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

logical-deductionmean-median-mode-rangesystematic-enumerationcasework caseworksystematic-enumeration ↑ Prerequisites: systematic-enumerationmean-median-mode-range

📏 Medium solution 💡 3 insights

Problem

The numbers $-2, 4, 6, 9$ and $12$ are rearranged according to these rules:

    1. The largest isn't first, but it is in one of the first three places. 
    2. The smallest isn't last, but it is in one of the last three places. 
    3. The median isn't first or last.

What is the average of the first and last numbers?

Pick an answer.

(A)

3.5

(B)

(C)

6.5

(D)

7.5

(E)

AMC 8 2004 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: The five numbers $-2, 4, 6, 9, 12$ are placed in a row of five positions, subject to three rules: (1) the largest is not first but is in one of the first three positions, (2) the smallest is not last but is in one of the last three positions, and (3) the median is not first and not last. Find the average of the numbers that end up in the first and last positions.

Givens: The five numbers to arrange: $-2, 4, 6, 9, 12$; Rule 1: largest ($12$) is not in position $1$, but is in position $1$, $2$, or $3$; Rule 2: smallest ($-2$) is not in position $5$, but is in position $3$, $4$, or $5$; Rule 3: median ($6$) is not in position $1$ or position $5$; Answer choices: (A) $3.5$, (B) $5$, (C) $6.5$, (D) $7.5$, (E) $8$

Unknowns: The average of the numbers in position $1$ and position $5$

Understand

Plan

Primary tool: #7 Break Into Subproblems

Secondary: #13 Count Smartly

Three independent rules pin down where three specific numbers can go, so Tool #7 (Break Into Subproblems) is the natural move: handle each rule on its own, then combine the results. After narrowing each of the three key numbers ($12$, $-2$, $6$) down to positions $\{2,3,4\}$, Tool #13 (Count Smartly) finishes the job: three distinct numbers must fill the three middle slots, leaving the remaining two numbers ($4$ and $9$) for positions $1$ and $5$.

Execute — Answer: C

#7 Break Into Subproblems 6.SP.B.5 Step 1

Identify the three special numbers.
Sort $-2, 4, 6, 9, 12$ to see that the largest is $12$, the smallest is $-2$, and the median (middle value) is $6$.
The other two numbers, $4$ and $9$, are not constrained by any rule.

$$\text{largest}=12,\ \ \text{smallest}=-2,\ \ \text{median}=6,\ \ \text{unconstrained}: 4, 9$$

💡 The Grade 6 "measures of center" idea labels min, max, and median. Naming them first turns each rule into a statement about a single number.

#7 Break Into Subproblems 6.EE.B.5 Step 2

Apply Rule 1 to the largest.
The number $12$ is in position $1$, $2$, or $3$ but not in position $1$, so $12$ must be in position $2$ or $3$.

$$\text{position of }12 \in \{1,2,3\} \cap \{2,3,4,5\} = \{2, 3\}$$

💡 Two constraints on the same variable become an intersection of the allowed sets. This is exactly the Grade 6 "which values make the statement true" move.

#7 Break Into Subproblems 6.EE.B.5 Step 3

Apply Rule 2 to the smallest.
The number $-2$ is in position $3$, $4$, or $5$ but not in position $5$, so $-2$ must be in position $3$ or $4$.

$$\text{position of }-2 \in \{3,4,5\} \cap \{1,2,3,4\} = \{3, 4\}$$

💡 Same intersection idea: "in the last three" combined with "not last" leaves only positions $3$ and $4$.

#7 Break Into Subproblems 6.EE.B.5 Step 4

Apply Rule 3 to the median.
The number $6$ cannot be in position $1$ or position $5$, so $6$ must be in position $2$, $3$, or $4$.

$$\text{position of }6 \in \{1,2,3,4,5\} \setminus \{1,5\} = \{2, 3, 4\}$$

💡 "Not first and not last" simply removes the two end positions from the choices.

#13 Count Smartly 6.EE.B.5 Step 5

Count the slots used by the three constrained numbers.
All three of $12$, $-2$, and $6$ are forced into the middle three positions $\{2,3,4\}$.
Three distinct numbers filling three slots means positions $2$, $3$, $4$ are taken in some order by exactly $\{-2, 6, 12\}$.

$$\{\text{positions of } 12, -2, 6\} = \{2,3,4\}$$

💡 Three items into three boxes is a pigeonhole-style count: every middle slot is used up by a special number, leaving the ends for the leftovers.

#13 Count Smartly 6.SP.B.5 Step 6

Identify the first and last numbers and compute their average.
With positions $2$, $3$, $4$ used up, the only numbers left for positions $1$ and $5$ are $4$ and $9$.
The average is their sum divided by $2$.

$$\text{average} = \dfrac{4 + 9}{2} = \dfrac{13}{2} = 6.5 \;\Rightarrow\; \textbf{(C)}$$

💡 Once the ends are pinned down to $\{4, 9\}$, the question reduces to a simple Grade 6 average.

[1] #7 6.SP.B.5 Identify the three special numbers. Sort $-2, 4, 6, 9, 12$ to see that the large

[2] #7 6.EE.B.5 Apply Rule 1 to the largest. The number $12$ is in position $1$, $2$, or $3$ but

[3] #7 6.EE.B.5 Apply Rule 2 to the smallest. The number $-2$ is in position $3$, $4$, or $5$ bu

[4] #7 6.EE.B.5 Apply Rule 3 to the median. The number $6$ cannot be in position $1$ or position

[5] #13 6.EE.B.5 Count the slots used by the three constrained numbers. All three of $12$, $-2$,

[6] #13 6.SP.B.5 Identify the first and last numbers and compute their average. With positions $2

Review

Reasonableness: Pick a valid arrangement and verify all three rules hold. Try $4,\ 12,\ -2,\ 6,\ 9$: position $1$ is $4$ (not $12$, good), $12$ is in position $2$ (one of the first three, good); $-2$ is in position $3$ (one of the last three, not last, good); $6$ is in position $4$ (not first, not last, good). All rules are satisfied, and the first and last numbers are $4$ and $9$, averaging $6.5$. The answer matches (C).

Alternative: Tool #14 (Extreme Cases): focus on the unconstrained numbers. The rules only mention the largest, smallest, and median. So $4$ and $9$ have no individual restrictions, and they are the only candidates that can land in position $1$ or position $5$ once the constrained numbers are forced into the middle. Their average $(4+9)/2 = 6.5$ pins down (C) without even checking specific orderings.

CCSS standards used (min grade 6)

6.SP.B.5 Summarize numerical data sets, including measures of center such as median (Identifying the smallest, largest, and median of $\{-2,4,6,9,12\}$, and computing the final average $(4+9)/2 = 6.5$.)
6.EE.B.5 Understand solving an equation or inequality as a process of answering which values make the statement true (Treating each rule as a constraint on a position, then intersecting the allowed positions for each of $12$, $-2$, and $6$.)

⭐ When a problem hands you several rules, work on one rule at a time. Each rule shrinks where a specific number can go; once three numbers are squeezed into three middle slots, the ends are forced — and that is all you need for the average.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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