AMC 8 · 2016 · #18
Easy mode Grade 4Problem
A 100-meter dash competition has 216 sprinters. The track has 6 lanes, so each race has exactly 6 runners.
After each race, the 5 runners who didn't win are out of the competition. Only the winner of each race moves on to race again later.
This keeps going until just one champion is left.
How many races are needed to find the champion?
Pick an answer.
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.
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
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) $72$
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
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.
💡 Shrinking $216$ down to $36$ keeps the structure (a power of $6$) but lets you count by hand.
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.
💡 The pattern from the easier problem generalizes: $5$ eliminations per race, no matter how big the field.
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.
💡 One division finishes the problem once the elimination rate is in hand.
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.
💡 Splitting the tournament into rounds — each its own counting subproblem — confirms the same total.
3.OA.A.3 Use Tool #9: try the smallest case where the track is exactly full. With $6$ spr 3.OA.B.5 Count eliminations to find the rule. In each small case, the champion is the onl 4.OA.A.3 Apply the rule to $216$ sprinters. Everyone except the champion must be eliminat 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.3Use 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.5Apply 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.3Solve 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.
⭐ Big tournament numbers shrink fast once you spot that every race eliminates exactly $5$ runners — Grade 4 division finishes it.
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