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 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.

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.

How to solve

Strategy Break Into Subproblems

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.

1STEP 1

Sort the list: largest 12, smallest -2, median 6 — the two leftovers 4 and 9 face no rule at all.

largest=12, smallest=-2, median=6, unconstrained: 4, 9

2STEP 2

Rule 1 puts 12 in the first three but not first, so 12 sits in position 2 or 3.

position of 12 ∈ {1,2,3} ∩ {2,3,4,5} = {2, 3}

3STEP 3

Rule 2 puts -2 in the last three but not last, so -2 sits in position 3 or 4.

position of -2 ∈ {3,4,5} ∩ {1,2,3,4} = {3, 4}

4STEP 4

Rule 3 keeps 6 off both ends, so the median 6 lands in position 2, 3, or 4.

position of 6 ∈ {1,2,3,4,5} ∖ {1,5} = {2, 3, 4}

5STEP 5

All three of 12, -2, 6 must sit in positions 2, 3, 4, so those three middle slots are completely filled.

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

6STEP 6

That leaves only 4 and 9 for the two ends, and their average is $\frac{4 + 9}{2}$ = 6.5 → (C).

average =

\frac{4 + 9}{2}

\frac{13}{2}

= 6.5 → (C)

Answer

6.5

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).

💡Key takeaway

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.

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