AMC 10 · 2023 · #16
Grade 7 arithmeticProblem
In a table tennis tournament, every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was 40% more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?
Pick an answer.
AMC 10 2023 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: In a round-robin tennis tournament every player meets every other player exactly once. There are twice as many right-handed players as left-handed players, and left-handers won $40\%$ more total games than right-handers. Find the total number of games played.
Givens: Round-robin: every pair of players plays once; Right-handed count is twice the left-handed count: $R = 2L$; Left-handers won $40\%$ more games than right-handers: $W_L = 1.4 \cdot W_R$; No ties and no ambidextrous players; Answer choices: (A) $15$, (B) $36$, (C) $45$, (D) $48$, (E) $66$
Unknowns: $G$, the total number of games played
Understand
Restated: In a round-robin tennis tournament every player meets every other player exactly once. There are twice as many right-handed players as left-handed players, and left-handers won $40\%$ more total games than right-handers. Find the total number of games played.
Givens: Round-robin: every pair of players plays once; Right-handed count is twice the left-handed count: $R = 2L$; Left-handers won $40\%$ more games than right-handers: $W_L = 1.4 \cdot W_R$; No ties and no ambidextrous players; Answer choices: (A) $15$, (B) $36$, (C) $45$, (D) $48$, (E) $66$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #7 Identify Subproblems
Multiple-choice with only five candidate values for $G$ is the textbook signal for Tool #3 (Eliminate Possibilities). We split the work with Tool #7: extract two clean tests that any valid $G$ must pass — (i) the $1.4$-ratio of wins forces $G$ to be a multiple of $12$, and (ii) round-robin forces $G$ to be a triangular number $\binom{N}{2}$. Then test each answer choice against both filters. The unique survivor is the answer.
Execute — Answer: B
6.RP.A.3 Step 1 - Subproblem A: derive the multiple-of-$12$ filter from the win ratio.
- Every game has one winner, so $G = W_R + W_L$.
- The ratio condition gives $W_L = \tfrac{7}{5} W_R$, so $G = W_R + \tfrac{7}{5} W_R = \tfrac{12}{5} W_R$.
- For $W_R$ to be an integer, $G$ must be a multiple of $12$.
💡 A clean way to read "$40\%$ more" is the ratio $7:5$; whenever a total splits in ratio $7:5$, the total must be a multiple of $12$.
7.SP.C.8 Step 2 - Subproblem B: derive the triangular-number filter from the round-robin structure.
- Each of the $N$ players plays every other player once, so $G = \binom{N}{2} = \tfrac{N(N-1)}{2}$ for some integer $N \ge 2$.
- So $G$ must be a triangular number.
💡 Round-robin = pick a pair. "Pick $2$ from $N$" is the triangular-number formula $\tfrac{N(N-1)}{2}$.
4.OA.B.4 Step 3 - Apply Filter A (multiple of $12$) to the five answer choices.
- $15, 45, 66$ are not multiples of $12$ and are eliminated.
- Only $36 = 12 \cdot 3$ and $48 = 12 \cdot 4$ survive.
💡 Checking divisibility by $12$ is a fast filter that knocks out three of the five options in one pass.
6.EE.B.6 Step 4 - Apply Filter B (triangular number) to the survivors.
- For $G = 36$: $\tfrac{N(N-1)}{2} = 36 \Rightarrow N(N-1) = 72 = 9 \cdot 8$, so $N = 9$ works.
- For $G = 48$: $N(N-1) = 96$, but $9 \cdot 10 = 90$ and $10 \cdot 11 = 110$, so no integer $N$ exists.
- Only $G = 36$ survives both filters.
💡 Two consecutive integers whose product is $72$ jump out as $8$ and $9$; $96$ has no such pair.
6.RP.A.3 Step 5 - Sanity check: with $G = 36$ and $N = 9$, the player split is $R = 6$ right-handed and $L = 3$ left-handed (since $N = 3L = 9$).
- Wins distribute as $W_R = \tfrac{5}{12} \cdot 36 = 15$ and $W_L = \tfrac{7}{12} \cdot 36 = 21$, and $21 / 15 = 1.4$ — the $40\%$ excess holds.
💡 If the survivor passes the original ratio condition exactly, the elimination is airtight.
6.RP.A.3 Subproblem A: derive the multiple-of-$12$ filter from the win ratio. Every game 7.SP.C.8 Subproblem B: derive the triangular-number filter from the round-robin structure 4.OA.B.4 Apply Filter A (multiple of $12$) to the five answer choices. $15, 45, 66$ are n 6.EE.B.6 Apply Filter B (triangular number) to the survivors. For $G = 36$: $\tfrac{N(N-1 6.RP.A.3 Sanity check: with $G = 36$ and $N = 9$, the player split is $R = 6$ right-hande Review
Reasonableness: Magnitude check: $9$ players in a round-robin play $\binom{9}{2} = 36$ games, which sits comfortably between the smaller (15, 36) and larger (48, 66) options. The win counts $15$ and $21$ are non-negative integers and sum to the total $36$ — internally consistent. The $1.4$ ratio reproduces "$40\%$ more" exactly. Every answer choice except $36$ fails one of the two independent filters, so the elimination is forced.
Alternative: Tool #6 (Guess and Check) reaching for Tool #13 (Convert to Algebra) at the end: let $L$ be the number of left-handers, so $R = 2L$ and $N = 3L$. Try $L = 1, 2, 3, 4$ and compute $G = \binom{3L}{2}$, then check whether $G$ is a multiple of $12$. $L = 1$: $G = 3$. $L = 2$: $G = 15$. $L = 3$: $G = 36$ — multiple of $12$, stop. Same answer (B) $36$.
CCSS standards used (min grade 7)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Turning "$40\%$ more" into the win ratio $W_L : W_R = 7 : 5$, then back-solving wins from the total.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Recognizing that a round-robin tournament with $N$ players plays $\binom{N}{2}$ games — the "pick $2$ from $N$" count.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Quickly testing each answer choice for divisibility by $12$ to eliminate three of the five options.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Treating $\binom{N}{2} = G$ as an equation in $N$ and reading off $N(N-1) = 2G$ to check whether each survivor is triangular.)
⭐ Two clean conditions on the total — multiple of $12$ from the win ratio, and a "$\binom{N}{2}$ pairs" count from the round-robin — knock out four of the five choices, leaving $\textbf{(B) }36$. When a multiple-choice problem hands you the answers, build small filters that each candidate must pass and the answer falls out.
⭐ Two clean conditions on the total — multiple of $12$ from the win ratio, and a "$\binom{N}{2}$ pairs" count from the round-robin — knock out four of the five choices, leaving $\textbf{(B) }36$. When a multiple-choice problem hands you the answers, build small filters that each candidate must pass and the answer falls out.
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