AMC 8 · 2012 · #22
Grade 6 arithmeticProblem
Let R be a set of nine distinct integers. Six of the elements are 2, 3, 4, 6, 9, and 14. What is the number of possible values of the median of R?
Pick an answer.
AMC 8 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.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: $R$ has exactly $9$ distinct integers; Six known elements: $2, 3, 4, 6, 9, 14$; Three unknown elements are free, as long as they are integers distinct from each other and from the known six; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Unknowns: The number of possible values of the median of $R$
Understand
Restated: 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.
Givens: $R$ has exactly $9$ distinct integers; Six known elements: $2, 3, 4, 6, 9, 14$; Three unknown elements are free, as long as they are integers distinct from each other and from the known six; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #2 Make a Systematic List
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 $5$th element after sorting, so the question reduces to: which positions in the sorted known list can become the $5$th slot when we slide three free integers into the line?
Execute — Answer: D
6.SP.B.5 Step 1 - Pin down what the median is.
- With $9$ distinct integers in ascending order, the median is the $5$th number.
- So we need to find every integer that can sit in the $5$th slot of some valid arrangement.
💡 Reducing "median of a 9-element set" to "the 5th smallest" is the Grade 6 definition of median for an odd-sized data set.
6.SP.B.5 Step 2 - Find the smallest possible median (an easier sub-question).
- Push all three unknown integers below $2$ so they fill slots $1, 2, 3$.
- The known list $2, 3, 4, 6, 9, 14$ then occupies slots $4$ through $9$.
- The $5$th slot is $3$.
💡 Solving the extreme "how small can it be?" first is the Tool #9 move — replace the general question with an easier boundary version.
6.SP.B.5 Step 3 - Find the largest possible median (the other extreme).
- Push all three unknowns above $14$ so they take slots $7, 8, 9$.
- The known list now sits in slots $1$ through $6$, and the $5$th slot is $9$.
💡 The companion extreme — "how large can it be?" — caps the range. Together the two extremes bound every possible median between $3$ and $9$.
6.SP.B.5 Step 4 - List the candidates between the extremes and confirm each is reachable.
- The possible medians are integers from $3$ to $9$: that is $\{3, 4, 5, 6, 7, 8, 9\}$.
- Build a witness arrangement for each value by choosing where to drop the three unknowns.
💡 A systematic list of one witness per value (Tool #2) shows nothing in $\{3, \ldots, 9\}$ gets skipped.
4.OA.A.3 Step 5 - Count the integers from $3$ to $9$ inclusive.
- That gives $9 - 3 + 1 = 7$ possible medians.
💡 Counting consecutive integers from $a$ to $b$ as $b - a + 1$ is a standard Grade 4 word-problem move.
6.SP.B.5 Pin down what the median is. With $9$ distinct integers in ascending order, the 6.SP.B.5 Find the smallest possible median (an easier sub-question). Push all three unkno 6.SP.B.5 Find the largest possible median (the other extreme). Push all three unknowns ab 6.SP.B.5 List the candidates between the extremes and confirm each is reachable. The poss 4.OA.A.3 Count the integers from $3$ to $9$ inclusive. That gives $9 - 3 + 1 = 7$ possibl Review
Reasonableness: The median of an odd-sized list sits at a fixed position ($5$th 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 $5$th-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.
Alternative: Tool #3 (Eliminate Possibilities) on the answer choices: the minimum median is clearly $\le 3$ (drop three tiny numbers below $2$) and the maximum is clearly $\ge 9$ (drop three huge numbers above $14$), so the count is at least $9 - 3 + 1 = 7$. The known set contains no gaps wider than $\{6,9\}$, but the unknowns can fill $7$ or $8$, so every integer in $[3,9]$ is reachable — exactly $7$. Choices (A), (B), (C) are too small and (E) overcounts.
CCSS standards used (min grade 6)
6.SP.B.5Summarize numerical data sets in relation to their context, including measures of center (mean, median, mode) (Using the Grade 6 definition of median — the middle value of an ordered data set — to identify the $5$th slot of $9$ sorted integers as the quantity we must control.)4.OA.A.3Solve multi-step word problems using the four operations (Counting the integers from $3$ to $9$ inclusive as $9 - 3 + 1 = 7$ at the final tally step.)
⭐ This AMC 8 problem only needs the Grade 6 definition of median — the middle number of a sorted list — that you already know!
⭐ This AMC 8 problem only needs the Grade 6 definition of median — the middle number of a sorted list — that you already know!
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