AMC 10 · 2012 · #15
Grade 5 arithmeticIn a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end of the tournament?
Pick an answer.
AMC 10 2012 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 each play every other team exactly once, and every game has one winner. After counting each team's wins, find the largest number of teams that could end up sharing the top win total.
Givens: There are $6$ teams; Every pair of teams plays exactly one game (round-robin); Each game produces exactly one winner and one loser (no ties in a game); Teams are ranked by number of games won; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Unknowns: The maximum possible number of teams that share the highest win count
Understand
Restated: Six teams each play every other team exactly once, and every game has one winner. After counting each team's wins, find the largest number of teams that could end up sharing the top win total.
Givens: There are $6$ teams; Every pair of teams plays exactly one game (round-robin); Each game produces exactly one winner and one loser (no ties in a game); Teams are ranked by number of games won; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Plan
Primary tool: #14 Extreme Principle
Secondary: #16 Change Focus / Count the Complement, #3 Eliminate Possibilities, #1 Draw a Diagram
The question asks for a maximum, so Tool #14 (Extreme Principle) drives it: push the number of tied teams as high as possible and test the boundary from the top down. Tool #16 (Change Focus) supplies the key fixed quantity — the total number of wins never changes, no matter who beats whom. Tool #3 (Eliminate Possibilities) kills the impossible top choice, and Tool #1 (Draw a Diagram) builds an actual win/loss table to prove the surviving count really happens.
Execute — Answer: D
4.NBT.B.5 Step 1 Count the total wins
- Every pair of the $6$ teams plays once.
- Each team faces $5$ others, giving $6 \times 5 = 30$ team-games, but that counts each game twice (once per team), so there are $30 \div 2 = 15$ games.
- Every game hands out exactly one win, so the six teams collect $15$ wins in total, always.
💡 One win is created per game, so the pile of wins is fixed at the number of games no matter who wins them.
5.NF.B.3 Step 2 Rule out all six tying
- If all $6$ teams tied, they would split the $15$ wins equally: $15 \div 6 = 2.5$ wins each.
- A team cannot win half a game, so an equal six-way split is impossible.
- Choice (E) $6$ is out.
💡 You can only tie if the fixed total splits evenly, and $15$ does not divide evenly among $6$.
4.OA.A.3 Step 3 Push the tie to five teams
- Try the next value down, $5$ tied teams.
- Give each of them $3$ wins: that uses $5 \times 3 = 15$ wins, exactly the whole pile.
- The leftover sixth team then has $15 - 15 = 0$ wins, which is below $3$, so those five really are the top.
- The numbers balance, so $5$ is possible in principle.
💡 The most teams can tie at the top is set by squeezing the win total to leave the odd team out with the fewest.
4.OA.C.5 Step 4 Build a table that works
- Make it real.
- Let Team $6$ lose every game; each of the other five then already has $1$ win from beating Team $6$.
- Among themselves those five play a round of $10$ games: number them $1$ to $5$ around a circle and let each team beat the next two clockwise.
- Every one of the five wins exactly $2$ of those games, for $1 + 2 = 3$ wins each, while Team $6$ stays at $0$.
- Five teams tie at $3$ wins, so the maximum is $5$, choice (D).
💡 A repeating "beat the next two" rule around a circle spreads the wins perfectly evenly among the five.
4.NBT.B.5 Every pair of the $6$ teams plays once. Each team faces $5$ others, giving $6 \t 5.NF.B.3 If all $6$ teams tied, they would split the $15$ wins equally: $15 \div 6 = 2.5$ 4.OA.A.3 Try the next value down, $5$ tied teams. Give each of them $3$ wins: that uses $ 4.OA.C.5 Make it real. Let Team $6$ lose every game; each of the other five then already Review
Reasonableness: The construction is consistent: $5$ teams with $3$ wins plus $1$ team with $0$ wins totals $5 \times 3 + 0 = 15$, matching the required $15$ wins exactly. The top total $3$ is above the leftover team's $0$, so the five genuinely share first place, and six was ruled out because $15$ is not a multiple of $6$. So $5$ is both possible and the largest workable value, confirming (D).
Alternative: Argue from the top choice down using elimination alone: $6$ fails since $15/6$ is not whole; for $5$, aim each top team at $3$ wins ($5 \times 3 = 15$) which forces the sixth team to $0$ — achievable — so no need to even check $4$, $3$, or $2$, because a larger tie has already been built.
CCSS standards used (min grade 5)
4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Counting games and wins: $6 \times 5 = 30$, then halving to $15$, and checking $5 \times 3 = 15$.)5.NF.B.3Interpret a fraction as division of the numerator by the denominator (Reading $15 \div 6 = 2.5$ as a non-whole share to prove a six-way tie is impossible.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Balancing the win total ($5 \times 3 = 15$, leaving $0$) to see that five tied teams fits the fixed count.)4.OA.C.5Generate a number or shape pattern following a given rule (Using the repeating "beat the next two around a circle" rule to give each of the five teams equal wins.)
⭐ The wins add up to a fixed number, so the most teams can tie at the top is however many still let that fixed total split evenly.
⭐ The wins add up to a fixed number, so the most teams can tie at the top is however many still let that fixed total split evenly.
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