AMC 8 · 2019 · #19

Easy mode Grade 4

Problem

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?

$\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

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

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

#7 Identify Subproblems 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?"

$$6 \text{ outside games} + 4 \text{ inside games} = 10 \text{ games per team}$$

💡 Breaking the $10$ games into two cleaner groups is Tool #7 — each piece is now a small, easy counting problem.

#7 Identify Subproblems 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.

$$6 \text{ wins} \times 3 \text{ pts/win} = 18 \text{ pts per top team from outside games}$$

💡 Multiplying $6 \times 3$ to count points from $6$ wins is a Grade 3 multiplication word-problem move.

#7 Identify Subproblems 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 $6$.

$$\dfrac{6 \text{ games} \times 3 \text{ pts}}{3 \text{ teams}} = 6 \text{ pts each (upper bound)}$$

💡 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.

#7 Identify Subproblems 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.

$$2 \times 3 + 2 \times 0 = 6 \text{ pts per top team from inside games}$$

💡 Solving a two-step word problem (count wins, then turn wins into points) is a Grade 3 four-operations skill.

#7 Identify Subproblems 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 $\textbf{(C) }24$.

$$18 + 6 = 24 \;\Rightarrow\; \textbf{(C)}$$

💡 Adding two two-digit numbers ($18+6$) is the Grade 2 fluency step that closes the argument.

#3 Eliminate Possibilities 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.

$$\text{Max total inside points} = 6 \times 3 = 18 < 24 = 3 \times 8 \;\Rightarrow\; 26 \text{ impossible}$$

💡 Multiple-choice elimination using a clean upper bound is Tool #3 — it pins the answer at $24$.

[1] #7 4.OA.A.3 Label the top three teams $A, B, C$ and the bottom three $D, E, F$. Each team pl

[2] #7 3.OA.A.3 Solve Subproblem 1. To maximize their scores, let $A, B, C$ each beat $D, E, F$

[3] #7 4.OA.A.3 Set up Subproblem 2. Inside the top group there are $\binom{3}{2} \times 2 = 6$

[4] #7 3.OA.D.8 Show $6$ points each is reachable. In each of the three pairs ($A\text{ vs }B$,

[5] #7 2.NBT.B.5 Add the two subproblem answers to get each top team's total: $18 + 6 = 24$ point

[6] #3 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.5 Fluently add and subtract within 100 (Adding the two subproblem totals $18 + 6 = 24$ to get each top team's final score.)
3.OA.A.3 Solve multiplication and division word problems within 100 (Computing $6 \times 3 = 18$ points from $6$ wins against the bottom three teams.)
3.OA.D.8 Solve 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.3 Solve 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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