AMC 10 · 2012 · #15

Grade 5 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

combinations-basic extremal-construction ↑ Prerequisites: combinations-basic

📏 Medium solution 💡 3 insights

Problem

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.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Extreme Principle

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.

1STEP 1

Count the total wins

Each of the 6 teams plays 5 others, but 6 × 5 = 30 counts every game twice, so there are 15 games — hence 15 wins in all.

(6 × 5)/2 = 15 games = 15 total wins

2STEP 2

Rule out all six tying

A six-way tie would give each team 15 ÷ 6 = 2.5 wins, and no team wins half a game, so (E) 6 is out.

15 ÷ 6 = 2.5 (not a whole number)

3STEP 3

Push the tie to five teams

Next value down: five teams at 3 wins each uses 5 × 3 = 15, the whole pile, leaving the sixth team 0 wins — safely below 3.

5 × 3 = 15, 15 - 15 = 0

4STEP 4

Build a table that works

Let Team 6 lose everything; seat the other five in a circle, each beating the next two, so all hit 1 + 2 = 3 wins — a five-way tie, (D).

1 (beat Team 6) + 2 (beat next two) = 3 wins → (D)

Answer

The construction is consistent: 5 teams with 3 wins plus 1 team with 0 wins totals 5 × 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).

💡Key takeaway

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.

Count the total wins
Rule out all six tying
Push the tie to five teams
Build a table that works

Count the total wins

Rule out all six tying

Push the tie to five teams

Build a table that works

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