Problem 24 · 2015 AMC 8
Hard
Number Theory
linear-diophantinemod-3
A baseball league consists of two four-team divisions. Each team plays every other team in its division N games. Each team plays every team in the other division M games with N > 2M and M > 4. Each team plays a 76 game schedule. How many games does a team play within its own division?
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Answer: B — 48 games.
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Hint 1 of 2
Count a team's opponents first: in a 4-team division it has 3 rivals (N games each), and the other division has 4 teams (M games each). That's the one equation: 3N + 4M = 76.
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Hint 2 of 2
One equation, two unknowns — the inequalities and whole-number-ness do the rest. Two squeezes work together: 'mod 3' pins M to one residue class, while N > 2M plus M > 4 traps M in a tiny range.
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Approach: build one Diophantine equation, then squeeze M with mod + bounds
- Opponents: 3 division rivals at N games and 4 cross-division teams at M games ⇒ 3N + 4M = 76.
- Mod 3: 3N vanishes, so 4M ≡ 76 ≡ 1, i.e. M ≡ 1 (mod 3) — M ∈ {7, 10, 13, …}.
- Bound it: N > 2M gives 76 = 3N + 4M > 10M, so M < 7.6; combined with M > 4, only M = 7 survives (and it fits the mod-3 class).
- Then N = (76 − 4·7)/3 = 48/3 = 16, so games within the division = 3N = 48.
- Why this transfers: one equation with two whole-number unknowns isn't stuck — mod arithmetic narrows to a residue class and inequalities crop it to a single value.
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