AMC 8 · 2012 · #14

Grade 5 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

combinations-basiclinear-equations-one-varsystematic-enumeration convert-to-algebraguess-and-check ↑ Prerequisites: multi-digit-arithmeticcombinations-basic

📏 Medium solution 💡 2 insights

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Problem

In the BIG N middle school football conference, every team plays every other team exactly once. A total of 21 games were played in the 2012 season. How many teams are in the conference?

Pick an answer.

(A)

$hspace{.05in}6$

(B)

$hspace{.05in}7$

(C)

$hspace{.05in}8$

(D)

$hspace{.05in}9$

(E)

$hspace{.05in}10$

AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

A game is a pair of teams, so the question is really "how many pairs can n teams make?" Tool #2 (Make a Systematic List) lets us count those pairs directly for small n — for 3 teams the pairs are AB, AC, BC (3 games); for 4 teams the pairs are AB, AC, AD, BC, BD, CD (6 games). Tool #5 (Look for a Pattern) turns those counts into the rule "new team n adds n-1 new games," giving the running totals 1, 3, 6, 10, 15, 21, …. Tool #6 (Guess and Check) then matches 21 to a choice without any algebra.

1STEP 1

Start small: 2 teams make 1 game; add team C, it plays A and B for 2 more — 3 games total. New teams meet every earlier team once.

n = 2 → 1 game; n = 3 → 1 + 2 = 3 games

2STEP 2

See the pattern: the n-th team adds n-1 new games, so n teams total the sum 1 + 2 + 3 + … + (n-1).

games(n) = 1 + 2 + 3 + … + (n-1)

3STEP 3

Extend the running totals 1, 3, 6, 10, 15 … until they reach 21, keeping a small (teams, games) table.

n = 2,3,4,5,6,7 → games = 1, 3, 6, 10, 15, 21

4STEP 4

Guess and check the choices: n = 6 gives 15, n = 7 gives 21, n = 8 gives 28 — only the n = 7 choice hits 21 exactly.

n = 6 → 15 (too small); n = 7 → 21 ✓; n = 8 → 28 (too big)

Answer

hspace{.05in}7

Sanity check by a different counting argument. Each of the 7 teams plays the other 6 teams once, which seems like 7 × 6 = 42 game-slots. But every game is counted twice (once for each team), so the real game count is 42 ÷ 2 = 21. That matches the given total exactly, confirming n = 7.

💡Key takeaway

You can solve this AMC 8 problem just by listing how many games show up when you add one team at a time — no algebra needed.

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