Competition · AMC preparation · step 4 of 4
AMC 8 · 2016 · #18
Grade 4 countingarithmeticPick an answer.
AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Try the smallest full track
Tool #9: try the smallest full-track cases — 6 sprinters need 1 race, and 36 sprinters give 6 + 1 = 7 races.
Shrinking 216 down to 36 keeps the structure (a power of 6) but lets you count by hand.
3.OA.A.3Solve An Easier Related ProblemCount eliminations per race
Count eliminations: for 36 sprinters, 35 lose over 7 races, and 7 × 5 = 35 — so every race removes exactly 5 sprinters.
The pattern from the easier problem generalizes: 5 eliminations per race, no matter how big the field.
3.OA.B.5Solve An Easier Related ProblemApply the rule to 216
Apply the rule to 216: everyone but the champion is out, so = = 43 races.
One division finishes the problem once the elimination rate is in hand.
The number of races needed is the whole field of 216 with the single champion set aside, then shared into equal groups of five.
▸ Why?
The meet ends when just one runner is left, so the number of runners that must be eliminated is the whole field of 216 minus that single survivor.
▸ Why?
The field splits with no overlap into the one champion who is never eliminated and everyone else who is, so taking the champion away from the whole leaves exactly the group that must be eliminated.
▸ Why?
Each race eliminates exactly five runners, and a runner once eliminated never races again, so the eliminated runners fall into equal groups of five with one group per race.
▸ Why?
Equal groups of five, one per race, mean the total eliminations are the number of races counted five at a time — that is, races times five.
▸ Why?
Since the total eliminations equal the number of races times five, undoing that multiplication by dividing the total by five gives back the number of races.
Simulate the rounds to check
Cross-check with Tool #7 by rounds: 216→36 races, 36→6 races, 6→1 race, so 36 + 6 + 1 = 43 → (C).
Splitting the tournament into rounds — each its own counting subproblem — confirms the same total.
4.OA.A.3Identify SubproblemsBig tournament numbers shrink fast once you spot that every race eliminates exactly 5 runners — Grade 4 division finishes it.
- Try the smallest full track
- Count eliminations per race
- Apply the rule to 216
- Simulate the rounds to check
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