AMC 8 · 2019 · #19
Easy mode Grade 4Problem
Imagine 6 teams in a tournament. Every team plays every other team exactly twice.
The scoring rule is simple. A win gives the team 3 points. A draw gives the team 1 point. A loss gives the team 0 points.
After all the games are over, the three teams with the highest total scores end up tied. That is, the top three teams all have the same number of points.
We want to find the largest score that each of these top three teams could have.
What is that largest possible score?
Pick an answer.
AMC 8 2019 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: Six teams play a round-robin in which every pair meets exactly twice. A win is worth $3$ points, a draw $1$ point, and a loss $0$ points. After all $30$ games are played, the three teams at the top of the standings end up with the same total number of points. What is the largest score those three top teams can share?
Givens: $6$ teams; each pair plays $2$ games (so $\binom{6}{2} \times 2 = 30$ games in all); Win $= 3$ pts, draw $= 1$ pt (for each team), loss $= 0$ pts; The top three teams finish tied in total points; Answer choices: (A) $22$, (B) $23$, (C) $24$, (D) $26$, (E) $30$
Unknowns: The greatest possible common total of the three top teams
Understand
Restated: Six teams play a round-robin in which every pair meets exactly twice. A win is worth $3$ points, a draw $1$ point, and a loss $0$ points. After all $30$ games are played, the three teams at the top of the standings end up with the same total number of points. What is the largest score those three top teams can share?
Givens: $6$ teams; each pair plays $2$ games (so $\binom{6}{2} \times 2 = 30$ games in all); Win $= 3$ pts, draw $= 1$ pt (for each team), loss $= 0$ pts; The top three teams finish tied in total points; Answer choices: (A) $22$, (B) $23$, (C) $24$, (D) $26$, (E) $30$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities, #6 Guess and Check
The cleanest move is to split the $10$ games each top team plays into two groups: (i) games against the three bottom teams (where the top team is free to win every time) and (ii) games against the other two top teams (where every point earned by one top team is a point denied to another). That is exactly Tool #7 (Identify Subproblems). After we compute the best score using this split, Tool #3 (Eliminate Possibilities) lets us check the leftover answer choices — especially $26$ and $30$ — to confirm $24$ is the maximum that is actually achievable.
Execute — Answer: C
4.OA.A.3 Step 1 - Label the top three teams $A, B, C$ and the bottom three $D, E, F$.
- Each team plays $10$ games: $6$ against the other group and $4$ against its own group.
- Subproblem 1 is "how many points can each top team earn from games against $D, E, F$?" Subproblem 2 is "how do $A, B, C$ split the points from the games they play against each other?"
💡 Breaking the $10$ games into two cleaner groups is Tool #7 — each piece is now a small, easy counting problem.
3.OA.A.3 Step 2 - Solve Subproblem 1.
- To maximize their scores, let $A, B, C$ each beat $D, E, F$ in both of their $2$ games — that is $6$ wins against the bottom three.
- Each top team therefore picks up $6 \times 3 = 18$ points from outside games, and these wins don't conflict with each other.
💡 Multiplying $6 \times 3$ to count points from $6$ wins is a Grade 3 multiplication word-problem move.
4.OA.A.3 Step 3 - Set up Subproblem 2.
- Inside the top group there are $\binom{3}{2} \times 2 = 6$ games.
- Each game gives out $3$ points if it has a winner ($3+0$) and only $2$ points if it is a draw ($1+1$).
- So the most points the top three can share from these games is $6 \times 3 = 18$, split equally three ways: $18 \div 3 = 6$ points each — but only if we can actually find an outcome where every top team gets exactly $6$.
💡 Comparing decisive games ($3$ pts handed out) versus draws ($2$ pts) shows draws waste points, so for the maximum we look for decisive games only.
3.OA.D.8 Step 4 - Show $6$ points each is reachable.
- In each of the three pairs ($A\text{ vs }B$, $A\text{ vs }C$, $B\text{ vs }C$) they play twice; let each team win exactly one of those two games.
- Then every top team has $2$ inside wins and $2$ inside losses, contributing $2 \times 3 + 2 \times 0 = 6$ points each — perfectly tied.
💡 Solving a two-step word problem (count wins, then turn wins into points) is a Grade 3 four-operations skill.
2.NBT.B.5 Step 5 - Add the two subproblem answers to get each top team's total: $18 + 6 = 24$ points.
- So $24$ is achievable and is the largest equal total — answer $\textbf{(C) }24$.
💡 Adding two two-digit numbers ($18+6$) is the Grade 2 fluency step that closes the argument.
4.OA.A.3 Step 6 - Verify by eliminating bigger choices (Tool #3).
- $\textbf{(E) }30$ would force each of $A, B, C$ to win every game including against each other — impossible, since when $A$ plays $B$ they cannot both win.
- $\textbf{(D) }26$ would need $26 - 18 = 8$ points from $4$ inside games per team, so $24$ inside-points total; but only $6$ inside games are played and they give out at most $6 \times 3 = 18$ points total.
- So $24$ is genuinely the max.
💡 Multiple-choice elimination using a clean upper bound is Tool #3 — it pins the answer at $24$.
4.OA.A.3 Label the top three teams $A, B, C$ and the bottom three $D, E, F$. Each team pl 3.OA.A.3 Solve Subproblem 1. To maximize their scores, let $A, B, C$ each beat $D, E, F$ 4.OA.A.3 Set up Subproblem 2. Inside the top group there are $\binom{3}{2} \times 2 = 6$ 3.OA.D.8 Show $6$ points each is reachable. In each of the three pairs ($A\text{ vs }B$, 2.NBT.B.5 Add the two subproblem answers to get each top team's total: $18 + 6 = 24$ point 4.OA.A.3 Verify by eliminating bigger choices (Tool #3). $\textbf{(E) }30$ would force ea Review
Reasonableness: Total points awarded across all $30$ games is at most $30 \times 3 = 90$ (only if every game is decisive). If the three top teams each had $24$, they share $72$ of those $90$ points and the three bottom teams share at most $90 - 72 = 18$ — which is easy ($D, E, F$ split their internal $6$ games, scoring $6$ points each, total $18$, while losing every game to the top). Numbers balance, so $24$ is consistent with the whole tournament accounting.
Alternative: Tool #6 (Guess and Check) directly on the answer choices: for each candidate total $T$, the inside-game points per top team are $T - 18$, so the three top teams together need $3(T-18)$ inside points. The available inside points are at most $6 \times 3 = 18$, giving $T \le 24$. Plugging in, $T = 24$ needs exactly $18$ inside points, achievable with the win-once-lose-once construction; $T = 23, 22$ are also achievable but smaller. So the maximum is $T = 24$.
CCSS standards used (min grade 4)
2.NBT.B.5Fluently add and subtract within 100 (Adding the two subproblem totals $18 + 6 = 24$ to get each top team's final score.)3.OA.A.3Solve multiplication and division word problems within 100 (Computing $6 \times 3 = 18$ points from $6$ wins against the bottom three teams.)3.OA.D.8Solve two-step word problems using four operations within 100 (Counting wins and losses inside the top group ($2$ wins, $2$ losses) and converting them to points ($2 \times 3 + 2 \times 0 = 6$).)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Organizing the $10$-game split, bounding the inside-game points at $6 \times 3 = 18$, and eliminating the $26$ and $30$ choices.)
⭐ This AMC 8 problem only needs Grade 4 multi-step word-problem reasoning you already know — split the games into two groups, multiply, and add!
⭐ This AMC 8 problem only needs Grade 4 multi-step word-problem reasoning you already know — split the games into two groups, multiply, and add!
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