Competition · AMC preparation · step 4 of 4

AMC 8 · 2006 · #20

Grade 6 counting
Archetype Pair-Sum Invariant └ Sub-archetype Universe Total Minus Known
combinations-basicsystematic-enumerationlogical-deduction identify-subproblemscomplementary-counting ↑ Prerequisites: combinations-basicmulti-digit-arithmetic
📏 Short solution 💡 2 insights
Problem
Six players play a round-robin singles tournament — each pair plays exactly one game, no ties. The win counts so far are Helen 4, Ines 3, Janet 2, Kendra 2, Lara 2. Find how many games Monica won.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
3
(E)
4

AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Find an Invariant

The key fact that pins down Monica's wins without knowing the matchup details is an invariant: in a no-tie round-robin, every game produces exactly one win, so the total wins across all players equals the total games played. Tool #11 (Find an Invariant) captures that. Tool #7 (Identify Subproblems) breaks the work into two small pieces — (1) count the total games, (2) subtract the five known win counts — so the arithmetic stays clean.

1STEP 1

Count the total games

Count the games: each of the six players meets the other five, which double-counts, so total = 6×52\frac{6 × 5}{2} = 15 games.

Total games = (6 × 5)/2 = 15
2STEP 2

Apply the win-total invariant

No ties means each game gives exactly one win, so all six win totals add up to the 15 games.

Helen + Ines + Janet + Kendra + Lara + Monica = 15
3STEP 3

Subtract the known wins

Add the five known wins: 4 + 3 + 2 + 2 + 2 = 13, so Monica's wins = 15 - 13 = 2, choice (C).

4 + 3 + 2 + 2 + 2 = 13, Monica's wins = 15 - 13 = 2 → (C)
Answer
2
Check the count of games a different way: pair every player with every other player. The pairs are (H,I), (H,J), (H,K), (H,L), (H,M), (I,J), (I,K), (I,L), (I,M), (J,K), (J,L), (J,M), (K,L), (K,M), (L,M) — exactly 15 pairs, matching 6×52\frac{6 × 5}{2} = 15. Then the wins 4 + 3 + 2 + 2 + 2 + 2 = 15 line up with the 15 games. Also, each player played 5 games, so each individual win count must be between 0 and 5 — Monica's 2 fits comfortably and is consistent with the other players' tallies.
💡Key takeaway

In a no-tie round-robin, every game adds exactly one win to the scoreboard — so the wins always sum to the number of games played. Once you count 15 games, Monica's wins are just 15 minus the others.

  • Count the total games
  • Apply the win-total invariant
  • Subtract the known wins

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