AMC 10 · 2023 · #16

Grade 7 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

combinations-basicratio-proportionlinear-equations-one-vartriangular-numbers caseworkidentify-subproblemsbound-inequality-then-enumerate ↑ Prerequisites: combinations-basicratio-proportion

📏 Medium solution 💡 3 insights

Problem

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.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

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 C(N, 2). Then test each answer choice against both filters. The unique survivor is the answer.

1STEP 1

Every game has one winner, so G = W_R + W_L; the 7:5 win ratio makes G = $\frac{12}{5}$ W_R, so G must be a multiple of 12.

W_L = 1.4 W_R =

\frac{7}{5}

W_R → G =

\frac{12}{5}

W_R → 12 ∣ G

2STEP 2

Each of the N players meets every other once, so G = C(N, 2) = $\frac{N(N-1)}{2}$ — the total must be a triangular number.

G = C(N, 2) =

\frac{(N(N-1))}{2}

3STEP 3

Filter A (multiple of 12) knocks out 15, 45, 66; only 36 and 48 survive.

{15, 36, 45, 48, 66} → {36, 48}

4STEP 4

Filter B (triangular): 36 = C(9, 2) since 9·8 = 72, but 96 has no consecutive-integer pair, so only 36 passes both.

36 = C(9, 2) ✓, 48 ≠ C(N, 2) for any N ×

5STEP 5

Check: G = 36 gives N = 9, R = 6, L = 3, W_R = 15, W_L = 21, and $\frac{21}{15}$ = 1.4 — the 40% excess holds, confirming 36.

R = 6, L = 3, W_R = 15, W_L = 21,

\frac{W_L}{W_R}

= 1.4 → (B) 36

Answer

Magnitude check: 9 players in a round-robin play C(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.

💡Key takeaway

Two clean conditions on the total — multiple of 12 from the win ratio, and a "C(N, 2) pairs" count from the round-robin — knock out four of the five choices, leaving (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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