AMC 8 · 2016 · #18

Grade 4 countingarithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Multi-Segment Path or Count Sum

systematic-enumerationpattern-recognitionmulti-digit-arithmetic easier-related-problemidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticpattern-recognition

📏 Short solution 💡 2 insights

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Problem

In an All-Area track meet, $216$ sprinters enter a $100-$ meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?

Pick an answer.

(A)

$mbox{ }36$

(B)

$mbox{ }42$

(C)

$mbox{ }43$

(D)

$mbox{ }60$

(E)

$mbox{ }72$

AMC 8 2016 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: A track meet has $216$ sprinters and a $6$-lane track. In each race $6$ runners compete; the $5$ non-winners are eliminated and the winner runs again later. How many races are needed before exactly one champion is left?

Givens: Starting field: $216$ sprinters; Each race holds exactly $6$ sprinters ($6$ lanes); Each race produces $1$ winner and eliminates $5$ losers; The single winner of a race continues to a later race; Answer choices: (A) $36$, (B) $42$, (C) $43$, (D) $60$, (E) $72$

Unknowns: The total number of races needed to crown one champion

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #7 Identify Subproblems

$216$ sprinters is a lot to picture, so Tool #9 (Easier Related Problem) starts with a tiny field — say $6$ or $36$ sprinters — to spot the underlying rule. The rule that pops out is simple: every race kills exactly $5$ sprinters, and we need to kill everyone except the champion. From there, Tool #7 (Identify Subproblems) gives a clean cross-check: split the tournament into rounds and count the races in each round separately, then add.

Execute — Answer: C

#9 Solve an Easier Related Problem 3.OA.A.3 Step 1

Use Tool #9: try the smallest case where the track is exactly full.
With $6$ sprinters, $1$ race decides the champion.
With $36$ sprinters, Round 1 has $36 \div 6 = 6$ races leaving $6$ winners, then Round 2 has $1$ race for the champion, giving $6 + 1 = 7$ races total.

$$36 \text{ sprinters} \Rightarrow 6 + 1 = 7 \text{ races}$$

💡 Shrinking $216$ down to $36$ keeps the structure (a power of $6$) but lets you count by hand.

#9 Solve an Easier Related Problem 3.OA.B.5 Step 2

Count eliminations to find the rule.
In each small case, the champion is the only sprinter not eliminated.
For $36$ sprinters, $35$ get eliminated and $7$ races happened, and $7 \times 5 = 35$.
So every race removes exactly $5$ sprinters — that is the rule that scales to any field size.

$$\text{races} \times 5 = \text{eliminations}$$

💡 The pattern from the easier problem generalizes: $5$ eliminations per race, no matter how big the field.

#9 Solve an Easier Related Problem 4.OA.A.3 Step 3

Apply the rule to $216$ sprinters.
Everyone except the champion must be eliminated, which is $216 - 1 = 215$ sprinters.
Divide by $5$ eliminations per race.

$$\dfrac{216 - 1}{5} = \dfrac{215}{5} = 43 \text{ races}$$

💡 One division finishes the problem once the elimination rate is in hand.

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

Cross-check with Tool #7 by simulating rounds.
Round 1: $216 \div 6 = 36$ races, $36$ winners advance.
Round 2: $36 \div 6 = 6$ races, $6$ winners advance.
Round 3: $6 \div 6 = 1$ race, $1$ champion.
Add the rounds.

$$36 + 6 + 1 = 43 \;\Rightarrow\; \textbf{(C)}$$

💡 Splitting the tournament into rounds — each its own counting subproblem — confirms the same total.

[1] #9 3.OA.A.3 Use Tool #9: try the smallest case where the track is exactly full. With $6$ spr

[2] #9 3.OA.B.5 Count eliminations to find the rule. In each small case, the champion is the onl

[3] #9 4.OA.A.3 Apply the rule to $216$ sprinters. Everyone except the champion must be eliminat

[4] #7 4.OA.A.3 Cross-check with Tool #7 by simulating rounds. Round 1: $216 \div 6 = 36$ races,

Review

Reasonableness: Two independent methods give the same answer of $43$, which sits between the trap answers (A) $36$ (only Round 1) and (D) $60$ (a careless $300 \div 5$). Sanity check the eliminations: $43$ races $\times 5$ losers each $= 215$ eliminated, leaving exactly $1$ champion out of $216$. The arithmetic closes perfectly.

Alternative: Tool #3 (Eliminate Possibilities) on the multiple choices: any valid answer must satisfy $\text{races} \times 5 = 215$, so $\text{races} = 43$. Only choice (C) works — (A) $36 \times 5 = 180$, (B) $42 \times 5 = 210$, (D) $60 \times 5 = 300$, (E) $72 \times 5 = 360$ all miss the target $215$.

CCSS standards used (min grade 4)

3.OA.A.3 Use multiplication and division within $100$ to solve word problems (Working the easier $36$-sprinter case by computing $36 \div 6 = 6$ races per round.)
3.OA.B.5 Apply properties of operations as strategies to multiply and divide (Recognizing the constant rate of $5$ eliminations per race ($\text{races} \times 5 = \text{eliminations}$) from the worked small cases.)
4.OA.A.3 Solve multistep word problems with whole numbers using the four operations (Combining subtraction ($216 - 1 = 215$) and division ($215 \div 5 = 43$), and adding the per-round race counts ($36 + 6 + 1 = 43$).)

⭐ Big tournament numbers shrink fast once you spot that every race eliminates exactly $5$ runners — Grade 4 division finishes it.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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