AMC 8 · 2014 · #24

Grade 6 arithmeticlogic

Archetype Arithmetic Expression Decomposition

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?

$\textbf{(A) }2.5\qquad\textbf{(B) }3.0\qquad\textbf{(C) }3.5\qquad\textbf{(D) }4.0\qquad \textbf{(E) }4.5$

Pick an answer.

(A)

2.5

(B)

3.0

(C)

3.5

(D)

4.0

(E)

4.5

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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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