AMC 8 · 2015 · #24

Grade 7 algebranumber-theory

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

linear-diophantinedivisibility-rulescombinations-basic convert-to-algebrabound-inequality-then-enumeratecasework ↑ Prerequisites: linear-equations-one-vardivisibility-rules

📏 Medium solution 💡 3 insights

Problem

A baseball league has two divisions of 4 teams each. Inside a division, every pair of teams plays N games against each other; across divisions, every pair plays M games. We're told N ＞ 2M, M ＞ 4, and each team plays a total of 76 games. How many of those 76 games are against teams in the same division?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Write an Equation

Tool #1 (Write an Equation) is the natural first move: count one team's opponents in each division and translate "76 games total" into a single linear equation in N and M. That equation alone has many solutions, so we then use Tool #12 (Use Cases) — combine the two inequalities (N ＞ 2M, M ＞ 4) with the requirement that N and M are integers to narrow M to a tiny list (M = 5, 6, 7) and test each case. Only one survives, giving the unique (N, M) pair that answers the question.

1STEP 1

One team meets 3 rivals in its own division (3N games) and 4 teams across (4M), so 3N + 4M = 76.

3N + 4M = 76

2STEP 2

Solve for N so the condition N ＞ 2M becomes a bound on M alone: N = $\frac{76 - 4M}{3}$ .

3N = 76 - 4M → N =

\frac{76 - 4M}{3}

3STEP 3

Substituting into N ＞ 2M gives M ＜ 7.6; with M ＞ 4 and M an integer, M ∈ {5, 6, 7}.

\frac{76 - 4M}{3}

＞ 2M → 76 - 4M ＞ 6M → 76 ＞ 10M → M ＜ 7.6. With M ＞ 4 and M an integer: M ∈ {5, 6, 7}.

4STEP 4

Only M = 7 makes 76 - 4M divisible by 3, so M = 5 and 6 fail and N = 16.

M = 5: 76 - 20 = 56, and

\frac{56}{3}

is not an integer. M = 6: 76 - 24 = 52, and

\frac{52}{3}

is not an integer. M = 7: 76 - 28 = 48, and

\frac{48}{3}

= 16. ✓

5STEP 5

Check 16 ＞ 2 × 7 holds, then the same-division total is 3N = 3 × 16 = 48.

N = 16, M = 7: N ＞ 2M ⇔ 16 ＞ 14 ✓. Intra-division games = 3N = 3 × 16 = 48 → (B).

Answer

Plug N = 16, M = 7 back into the full season: 3(16) + 4(7) = 48 + 28 = 76. ✓ Both inequalities hold: 16 ＞ 2 · 7 = 14 and 7 ＞ 4. The split 48 intra + 28 inter = 76 matches the answer choice (B), and the intra-division share is more than half the schedule, which fits "N ＞ 2M" (you play same-division opponents more often than cross-division ones).

💡Key takeaway

Write the count as one equation, then let the inequalities and the "games must be whole numbers" rule shrink the choices until just one (N, M) pair is left.

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