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AMC 8 · 2006 · #20

Grade 6 counting
Archetype Pair-Sum Invariant
combinations-basicsystematic-enumerationlogical-deduction identify-subproblemscomplementary-counting ↑ Prerequisites: combinations-basicmulti-digit-arithmetic
📏 Short solution 💡 2 insights

Problem

A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 4 games, Ines won 3 games, Janet won 2 games, Kendra won 2 games and Lara won 2 games, how many games did Monica win?

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
3
(E)
4
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Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: There are $6$ players, each pair plays exactly one game; Every game has a winner and a loser (no ties); Wins: Helen $= 4$, Ines $= 3$, Janet $= 2$, Kendra $= 2$, Lara $= 2$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$

Unknowns: The number of games Monica won

Understand

Restated: 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.

Givens: There are $6$ players, each pair plays exactly one game; Every game has a winner and a loser (no ties); Wins: Helen $= 4$, Ines $= 3$, Janet $= 2$, Kendra $= 2$, Lara $= 2$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$

Plan

Primary tool: #11 Find an Invariant

Secondary: #7 Identify Subproblems

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.

Execute — Answer: C

#7 Identify Subproblems 5.NBT.B.5 Step 1
  • Count the total number of games.
  • Each of the $6$ players plays the other $5$, which counts every game twice (once from each player's side).
  • Divide by $2$ to fix the double count.
$$\text{Total games} \;=\; \dfrac{6 \times 5}{2} \;=\; 15$$

💡 "$6$ players, each plays $5$ others" sounds like $30$, but every game gets counted from both sides — halving fixes it.

#11 Find an Invariant 6.EE.A.2 Step 2
  • Apply the invariant.
  • Every game adds exactly $1$ win to somebody's tally, so the sum of all $6$ players' wins equals the number of games.
  • With $15$ games, the wins must total $15$.
$$\text{Helen} + \text{Ines} + \text{Janet} + \text{Kendra} + \text{Lara} + \text{Monica} \;=\; 15$$

💡 No ties means every game has exactly one winner, so total wins is locked equal to total games — no matter who beat whom.

#7 Identify Subproblems 6.EE.B.7 Step 3

Add the five known win counts and subtract from $15$ to isolate Monica's wins.

$$4 + 3 + 2 + 2 + 2 \;=\; 13,\quad \text{Monica's wins} \;=\; 15 - 13 \;=\; 2 \;\Rightarrow\; \textbf{(C)}$$

💡 The invariant turns the question into a single subtraction: total minus the part we know gives the part we want.

[1] #7 5.NBT.B.5 Count the total number of games. Each of the $6$ players plays the other $5$, wh
[2] #11 6.EE.A.2 Apply the invariant. Every game adds exactly $1$ win to somebody's tally, so the
[3] #7 6.EE.B.7 Add the five known win counts and subtract from $15$ to isolate Monica's wins.

Review

Reasonableness: 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 $\tfrac{6 \times 5}{2} = 15$. 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.

Alternative: Tool #13 (Convert to Algebra): let $M$ be Monica's wins. Total wins equals total games, so $4 + 3 + 2 + 2 + 2 + M = 15$, giving $13 + M = 15$ and $M = 2$. Same answer (C) — algebra just names the unknown and solves a one-step equation, which is what the invariant approach already does informally.

CCSS standards used (min grade 6)

  • 5.NBT.B.5 Fluently multiply multi-digit whole numbers using the standard algorithm (Computing the total number of games as $\tfrac{6 \times 5}{2} = 15$ from the round-robin pairing count.)
  • 6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Writing the invariant "sum of all six players' wins equals total games" as an expression that includes Monica's unknown win count.)
  • 6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ (Solving $13 + M = 15$ to get Monica's wins $M = 2$.)

⭐ 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.

⭐ 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.

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