Competition · AMC preparation · step 4 of 4

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

Identify the three special numbers

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

Apply the largest-number rule

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

Apply the smallest-number rule

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

Apply the median rule

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

Count the slots used

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}

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.

The largest, the smallest, and the median are all forced into positions 2, 3, and 4, so the two numbers that no rule mentions are the only ones left for the first and last positions.

▸ Why?

Each of the three special numbers is kept away from both end positions by its own rule, so all three can sit only somewhere among positions 2, 3, and 4.

▸ Why?

The largest is allowed only the first three positions and then barred from the very first, and the smallest is allowed only the last three and then barred from the very last, so taking away the barred end leaves each of them with positions inside the middle.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

The median is barred from the first and the last, and taking those two ends away from the whole row of five positions leaves exactly the middle three positions.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

There are exactly three special numbers and exactly three middle positions, and each number needs its own separate position, so the three numbers use up all three middle positions with none left over.

▸ Why?

Matching each of the three numbers to a separate one of the three middle positions is a one-to-one pairing, and a one-to-one pairing between two groups means they are the same size with nothing left unmatched.

One-to-one correspondenceIf each thing here pairs with exactly one thing there, the two groups are the same size.

▸ Why?

The row of five positions is the three middle positions together with the two ends, and the five numbers are the three special ones together with the other two, so once the special numbers fill the middle the remaining two numbers must fill the ends.

▸ Why?

The row splits into the middle three and the two ends with no gap and no overlap, so removing the part already filled leaves exactly the two end positions for the two remaining numbers.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

6.EE.B.5Convert To Algebra

6STEP 6

Average the end numbers

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

average = (4 + 9)/2 = 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.

Identify the three special numbers
Apply the largest-number rule
Apply the smallest-number rule
Apply the median rule
Count the slots used
Average the end numbers

A parent dashboard for the family lives at sensimlab.com.

Identify the three special numbers

Apply the largest-number rule

Apply the smallest-number rule

Apply the median rule

Count the slots used

Average the end numbers

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