Competition · AMC preparation · step 4 of 4
AMC 8 · 2012 · #14
Grade 5 countingPick an answer.
AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Start with the smallest cases
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.
Listing pairs in order (A first, then B, then C) is the systematic-list move — no pair gets missed or double-counted.
4.OA.A.3Make A Systematic ListSpot the pattern
See the pattern: the n-th team adds n-1 new games, so n teams total the sum 1 + 2 + 3 + … + (n-1).
Writing the count as a numerical expression (a sum of consecutive whole numbers) is a Grade 5 "express a calculation as an expression" move.
The total number of games in a round-robin among n teams equals the sum 1 + 2 + 3 + … + (n-1).
▸ Why?
Build the conference up one team at a time; when a team joins the k teams already present, it plays each of those earlier teams exactly once, so it brings in exactly k new games — as many new games as there were teams before it.
▸ Why?
Each new game the joining team plays is matched to exactly one earlier team, and each earlier team gives it exactly one such game, so the number of new games is the same as the number of earlier teams.
▸ Why?
Every game belongs to exactly one team — the later joiner of its two teams — so sorting all games this way splits them into separate groups of sizes 1, 2, 3, …, (n-1) with none left out and none counted twice, and the whole count is those group sizes added together.
Extend the running totals to 21
Extend the running totals 1, 3, 6, 10, 15 … until they reach 21, keeping a small (teams, games) table.
Each row adds the next whole number: 10 + 5 = 15, then 15 + 6 = 21 — there is the 21 we need.
4.OA.A.3Make A Systematic ListCheck the answer choices
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.
Choice (B) n = 7 is the only one that produces exactly 21 games — the answer is (B).
4.OA.A.3Guess And CheckYou 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.
- Start with the smallest cases
- Spot the pattern
- Extend the running totals to 21
- Check the answer choices
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