Competition · AMC preparation · step 4 of 4

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

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 c₁ ≤ c₂ ≤ … ≤ c₁₀₀. The median is the average of c₅₀ and c₅₁. What is the largest value this median can take?

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.

How to solve

Strategy Change Focus / Count the Complement

We want c₅₀ and c₅₁ 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 c₅₀, c₅₁ 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 (c₅₀, c₅₁), 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.

1STEP 1

Split the list into three blocks

Split the sorted list into three blocks: the 49 low customers, the median pair (c₅₀, c₅₁), and the 49 high customers, summing to 252.

c₁ + … + c₄₉_low (49) + c₅₀ + c₅₁_median pair + c₅₂ + … + c₁₀₀_high (49) = 252

2STEP 2

Minimize the low block

Give each of the 49 low customers the minimum 1 can, so the low block totals 49 cans and leaves the rest for the middle.

c₁ = c₂ = … = c₄₉ = 1, sum = 49 × 1 = 49

3STEP 3

Subtract off the low block

Subtract the low block from 252 to leave 203 cans for the median pair plus the high block (51 customers).

c₅₀ + c₅₁ + … + c₁₀₀ = 252 - 49 = 203

4STEP 4

Divide the rest among 51

Set all high values equal to c₅₁ and share 203 among 51 as evenly as possible: 203 = 50 × 4 + 3, so 50 get 4 cans and one gets 3.

203 ÷ 51 = 3 remainder 50, so 203 = 50 × 4 + 1 × 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."

Sharing the 203 cans that remain among the top 51 customers as evenly as the cans allow lifts the middle pair as high as it can go, which hands the fiftieth customer 3 cans and every customer from the fifty-first up 4 cans.

▸ Why?

The fiftieth customer's count cannot reach 4, because in the sorted line every one of the 51 customers from the fiftieth onward holds at least as many cans as the fiftieth, so 4 apiece would demand 204 cans while only 203 are left.

▸ Why?

If the fiftieth held at least 4 cans, then every customer from the fifty-first up holds at least as many as the fiftieth, so each of them holds at least 4 too, since a sorted list never dips as you move to the right.

Transitivity of orderIf the first is greater than the second and the second greater than the third, the first is greater than the third.

▸ Why?

51 customers each holding at least 4 cans need at least 51 fours in total, which comes to 204 cans.

▸ Why?

51 equal groups of 4 cans is 51 times 4, and equal groups pile up by multiplication to 204.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

▸ Why?

the block's cans are just the 51 separate counts added together, so if every count is at least 4 their total is at least 204.

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?

Giving 3 cans to the fiftieth customer and 4 to each of the other 50 in the top block is a real, valid arrangement, because those counts add back to exactly the 203 cans on hand and still rise from left to right (1, then 3, then 4).

▸ Why?

50 customers with 4 cans is 50 fours, or 200 cans, and one more customer with 3 makes 203 in all.

▸ Why?

50 equal groups of 4 cans total 50 times 4, which is 200 cans.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

▸ Why?

the top block's cans are separate parts of one total, so 200 and 3 combine to the full 203.

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

5.NBT.B.6Change Focus Count The Complement

5STEP 5

Average the middle two

The lone 3 must sit at c₅₀ to keep the order, giving c₅₀ = 3 and c₅₁ = 4, so the median is $\frac{3 + 4}{2}$ .

c₅₀ = 3, c₅₁ = 4, median = (3 + 4)/2 = 3.5 → (C)

Answer

3.5

Sanity check: the average is

\frac{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 × 1.52 ≈ 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

\frac{203}{51}

≈ 3.98 for the upper block, is consistent. Trying to push higher fails: if c₅₀ ≥ 4, the upper 51 values would need to sum to at least 4 × 51 = 204 ＞ 203.

💡Key takeaway

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!

Split the list into three blocks
Minimize the low block
Subtract off the low block
Divide the rest among 51
Average the middle two

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Split the list into three blocks

Minimize the low block

Subtract off the low block

Divide the rest among 51

Average the middle two

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