Competition · AMC preparation · step 4 of 4

AMC 8 · 2012 · #22

Grade 6 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
mean-median-mode-rangesystematic-enumerationbound-inequality-then-enumerate bound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: mean-median-mode-range
📏 Medium solution 💡 3 insights
Problem
A set R has 9 distinct integers. Six of them are {2, 3, 4, 6, 9, 14}; the other three are unknown. We must count how many different values the median of R can take.

Pick an answer.

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

AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Related Problem

Instead of analyzing all ways to pick the three unknown integers, solve the easier extreme cases first (Tool #9): What is the smallest median we can force? What is the largest? Once we have the range, use Tool #2 to systematically check each integer in that range and confirm it is reachable. The median of 9 numbers is always the 5th element after sorting, so the question reduces to: which positions in the sorted known list can become the 5th slot when we slide three free integers into the line?

1STEP 1

Pin down which value is the median

With 9 distinct integers in ascending order the median is the 5th number, so we hunt for every integer that can occupy that slot.

median = a₅ where a₁ ＜ a₂ ＜ … ＜ a₉
2STEP 2

Find the smallest possible median

Slide all three unknowns below 2 so the known six shift up; the 5th slot then lands on 3 — the smallest possible median.

x, y, z_ ＜ 2, 2, 3, 4, 6, 9, 14 → median = 3
3STEP 3

Find the largest possible median

Slide all three unknowns above 14 so the known six drop down; the 5th slot then lands on 9 — the largest possible median.

2, 3, 4, 6, 9, 14, x, y, z_ ＞ 14 → median = 9
4STEP 4

Check every value in between

Between the extremes every integer works: give each value in 3 to 9 a witness arrangement by dropping the unknowns into the right gaps.

3: {-2,-1,0}∪known → -2,-1,0,2,3,4,6,9,14 ; 4: {-1,0,1} → -1,0,1,3,4,6,9,14,? (use 14) ; 5: {0,1,5} → 0,1,2,4,5,6,9,14,… (insert 5 between 4 and 6) ; 6: {0,1,7} → 0,1,2,4,6,7,9,14,… ; 7: {7,10,11} → 2,3,4,6,7,9,10,11,14 ; 8: {8,10,11} → 2,3,4,6,8,9,10,11,14 ; 9: {15,16,17} → 2,3,4,6,9,14,15,16,17
5STEP 5

Count from 3 to 9

Count the integers from 3 to 9 inclusive: 9 - 3 + 1 = 7 possible medians.

9 - 3 + 1 = 7 → (D)
Answer
7
The median of an odd-sized list sits at a fixed position (5th of 9). The known six numbers {2,3,4,6,9,14} already span a wide range, so the three free integers act as sliders that shift the 5th-slot value from 3 (push them all left) up to 9 (push them all right). A range of 7 consecutive integers is consistent with three free slots pushing the median past 3, 4, 5, 6, 7, 8, 9 — answer (D) lands inside the choice list and matches the boundary calculation.
💡Key takeaway

This AMC 8 problem only needs the Grade 6 definition of median — the middle number of a sorted list — that you already know!

  • Pin down which value is the median
  • Find the smallest possible median
  • Find the largest possible median
  • Check every value in between
  • Count from 3 to 9

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