AMC 8 · 2019 · #19

• 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.

• 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.

• 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.

• 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.

• 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.

• 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 $24$.