AMC 8 · 2014 · #24

Grade 6 arithmeticlogic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

mean-median-mode-rangeoptimization-countingbound-inequality-then-enumerate bound-inequality-then-enumerateidentify-subproblems ↑ Prerequisites: mean-median-mode-rangemulti-digit-arithmetic

📏 Long solution 💡 4 insights

Problem

One day the Beverage Barn sold $252$ cans of soda to $100$ customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?

Pick an answer.

(A)

2.5

(B)

3.0

(C)

3.5

(D)

4.0

(E)

4.5

AMC 8 2014 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: On one day, $100$ customers bought a total of $252$ cans of soda, and every customer bought at least one can. Arrange the per-customer can counts in non-decreasing order and call them $c_1 \le c_2 \le \ldots \le c_{100}$. The median is the average of $c_{50}$ and $c_{51}$. What is the largest value this median can take?

Givens: There are $100$ customers, so the median is $\dfrac{c_{50} + c_{51}}{2}$ in the sorted list; Total cans sold $= 252$, so $c_1 + c_2 + \cdots + c_{100} = 252$; Every customer bought at least $1$ can: $c_i \ge 1$ for all $i$; Answer choices: (A) $2.5$, (B) $3.0$, (C) $3.5$, (D) $4.0$, (E) $4.5$

Unknowns: The maximum possible value of the median $\dfrac{c_{50} + c_{51}}{2}$

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem

We want $c_{50}$ and $c_{51}$ to be as large as possible, but the budget of $252$ cans is shared among all $100$ customers. Tool #16 (Change Focus) flips the question: instead of "how large can $c_{50}, c_{51}$ be?", ask "how small can everyone *else* be?" — the cans we save by minimizing the rest are exactly the cans we can hand to the median pair. Tool #7 (Subproblems) splits the sorted list into three blocks — the $49$ customers below the median, the median pair $(c_{50}, c_{51})$, and the $49$ customers above — each handled separately. Tool #9 (Easier Problem) checks the logic on a tiny version (say $6$ customers, $14$ cans) before trusting it on $100$.

Execute — Answer: C

#7 Identify Subproblems 6.SP.B.5 Step 1

Split the sorted list into three blocks: the $49$ "low" customers ($c_1$ to $c_{49}$), the median pair ($c_{50}, c_{51}$), and the $49$ "high" customers ($c_{52}$ to $c_{100}$).
The total $252$ is the sum across all three blocks.

$$\underbrace{c_1 + \cdots + c_{49}}_{\text{low (49)}} + \underbrace{c_{50} + c_{51}}_{\text{median pair}} + \underbrace{c_{52} + \cdots + c_{100}}_{\text{high (49)}} = 252$$

💡 Splitting at the median is a Grade 6 statistics move — the median is defined by its position in the sorted list, so the natural subproblems are the parts before, at, and after that position.

#16 Change Focus / Count the Complement 4.OA.A.3 Step 2

Apply Tool #16 to the low block.
To free up the most cans for the median pair, minimize the $49$ low values.
Since every $c_i \ge 1$, the smallest possible value for each is $1$, so the low block uses $49 \times 1 = 49$ cans.

$$c_1 = c_2 = \cdots = c_{49} = 1, \quad \text{sum} = 49 \times 1 = 49$$

💡 "To make the middle bigger, push the bottom as low as it can go" — a Grade 4 multi-step word-problem reasoning step (find the smallest allowed value, then add).

#7 Identify Subproblems 4.OA.A.3 Step 3

Subtract the low block from the total to see what is left for the median pair plus the high block ($51$ customers in total).

$$c_{50} + c_{51} + \cdots + c_{100} = 252 - 49 = 203$$

💡 After locking in the low block, the remaining cans must cover the other $51$ customers — a clean subtraction step from the subproblem split.

#16 Change Focus / Count the Complement 5.NBT.B.6 Step 4

Now apply Tool #16 again, but flipped, to the high block.
Because the list is sorted, every customer above the median satisfies $c_i \ge c_{51}$.
To keep $c_{51}$ as large as possible we should not "waste" cans pushing the high block higher than $c_{51}$ — set $c_{52} = c_{53} = \cdots = c_{100} = c_{51}$.
So the $51$ remaining customers will share $203$ cans as evenly as possible, with $c_{50}$ allowed to be one less than the rest if needed.

$$203 \div 51 = 3 \text{ remainder } 50, \quad \text{so } 203 = 50 \times 4 + 1 \times 3$$

💡 Dividing $203$ by $51$ with remainder is a Grade 5 long-division skill — the quotient $3$ and remainder $50$ tell us "$50$ customers get one extra can."

#16 Change Focus / Count the Complement 6.SP.B.5 Step 5

Assign cans: $50$ customers in the top $51$ get $4$ cans, and the remaining $1$ gets $3$ cans.
To keep the sort order, the lone $3$ must sit at the lowest position of this block — that is, at $c_{50}$.
So $c_{50} = 3$ and $c_{51} = c_{52} = \cdots = c_{100} = 4$.
Check: $c_{49} = 1 \le c_{50} = 3 \le c_{51} = 4$.

$$c_{50} = 3, \quad c_{51} = 4, \quad \text{median} = \dfrac{3 + 4}{2} = 3.5 \;\Rightarrow\; \textbf{(C)}$$

💡 Computing the median of an even-sized dataset as the average of the two middle values is the core Grade 6 statistics standard.

[1] #7 6.SP.B.5 Split the sorted list into three blocks: the $49$ "low" customers ($c_1$ to $c_{

[2] #16 4.OA.A.3 Apply Tool #16 to the low block. To free up the most cans for the median pair, m

[3] #7 4.OA.A.3 Subtract the low block from the total to see what is left for the median pair pl

[4] #16 5.NBT.B.6 Now apply Tool #16 again, but flipped, to the high block. Because the list is so

[5] #16 6.SP.B.5 Assign cans: $50$ customers in the top $51$ get $4$ cans, and the remaining $1$

Review

Reasonableness: Sanity check: the average is $252 / 100 = 2.52$. The median can exceed the mean only when the bottom half is squashed down and the top half is leveled, which is exactly what we did. Pushing the low $49$ to $1$ each saved $49 \times 1.52 \approx 75$ cans, which we redistributed to lift the upper $51$ from their fair share of $2.52$ up to nearly $4$. A median of $3.5$, slightly above the lifted average of $203/51 \approx 3.98$ for the upper block, is consistent. Trying to push higher fails: if $c_{50} \ge 4$, the upper $51$ values would need to sum to at least $4 \times 51 = 204 > 203$.

Alternative: Tool #9 (Easier Problem) on a baby version: $6$ customers, $14$ cans, each $\ge 1$, maximize the median (= avg of $c_3, c_4$). Minimize the bottom $2$: $c_1 = c_2 = 1$, leaving $12$ for $4$ customers. $12 \div 4 = 3$ evenly, so $c_3 = c_4 = c_5 = c_6 = 3$ and the median is $3$. The procedure matches — push the bottom to $1$, then split the rest as evenly as possible — confirming the strategy used on the full $100$-customer version.

CCSS standards used (min grade 6)

6.SP.B.5 Summarize numerical data sets in relation to their context (including measures of center) (Setting up the median of $100$ values as $\dfrac{c_{50} + c_{51}}{2}$ and computing the final answer $\dfrac{3 + 4}{2} = 3.5$.)
4.OA.A.3 Solve multi-step word problems with whole numbers, using the four operations (Minimizing the bottom $49$ customers to $1$ each (sum $= 49$) and subtracting from the total ($252 - 49 = 203$).)
5.NBT.B.6 Find whole-number quotients with up to four-digit dividends and two-digit divisors (Dividing $203 \div 51 = 3$ remainder $50$ to figure out how to split the $203$ remaining cans among $51$ customers.)

⭐ To make the middle of a sorted list as big as possible, squash the bottom as small as the rules allow — that frees up the most resources to lift the middle!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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