AMC 10 Step by Step
Printable problems

Games & Processes

Turn-based games, iterated processes, tournaments, scheduling

What you need to know

  • Round-robin. If nn teams each play every other team once, there are (n2)\binom{n}{2} games. With no ties, each game has one winner, so the wins also add up to (n2)\binom{n}{2}. Example: 88 teams play (82)=28\binom82 = 28 games and win 2828 in all.
  • Single elimination. nn players need exactly n−1n-1 matches: each match knocks out one player, and everyone but the champion is knocked out. Example: 6464 players need 6363 matches.
  • Invariants. An invariant is something no step changes, like even/odd, or a total's remainder when divided by some mm. A final state with a different value can never be reached. Example: 1,2,…,101, 2, \dots, 10 are on a board. A move erases two numbers and writes their difference, which lowers the sum by twice the smaller one. So the sum, 5555 at the start, stays odd, and the last number left is never 00.
  • Winning and losing positions. A position is losing if the player to move loses with best play on both sides. That happens when every move leads to a winning position. A position is winning when some move leads to a losing one. In "take 11 to kk stones; whoever takes the last stone wins," the losing positions are the multiples of k+1k+1. Example: taking 11 to 33 stones, 2020 is losing. Whatever you take, your opponent tops the round up to 44 stones, and in the end takes the last one.

How AMC 10 tests it

  • Problems 5–12: games in a tournament, handshakes at a party, days until a schedule repeats.
  • Problems 12–20: a number is changed again and again by a rule; find its value after many steps.
  • Problems 15–22: two-player games of taking objects away, asking which starting sizes are first-player wins.
  • Problems 18–25: a repeated process with a counting answer.

Standard approaches

  1. Work out the first several steps by hand; a cycle or pattern usually shows up fast.
  2. Look for an invariant the process keeps (even/odd, a sum's remainder).
  3. For two-player games, work backward from the end, marking positions W (winning) or L (losing).
  4. For tournaments, use total wins =(n2)= \binom{n}{2} to set up an equation.
  5. For "how many starting values," find the pattern, then count with remainders.

Worked example

Alice and Bob alternately remove 11, 33, or 44 coins from a pile of nn coins, Alice first; whoever takes the last coin wins. For how many integers nn with 1≤n≤1001 \le n \le 100 does Bob have a winning strategy?

(A) 1414 (B) 2020 (C) 2828 (D) 2929 (E) 3434

Solution.

Step 1: the rule. Mark each pile size L (losing) or W (winning) for the player about to move. A pile of 00 is L (the other player just took the last coin). A pile is W exactly when taking 11, 33, or 44 coins can leave an L pile.

Step 2: the table. For example, 22 is L (its only move leaves 11, a W), and 77 is L (its moves leave 66, 44, 33, all W).

nn001122334455667788991010111112121313141415151616
W/LLWLWWWWLWLWWWWLWL

Step 3: the period. Each entry depends only on the entries 11, 33, and 44 places back. Entries 77 to 1010 (L, W, L, W) match entries 00 to 33, so from there on the table repeats every 77. The L piles are the numbers with remainder 00 or 22 when divided by 77 (written n≡0n \equiv 0 or 2(mod7)2 \pmod 7).

Step 4: who wins. Alice moves first, so Bob wins exactly when the starting pile is L.

Step 5: count. Multiples of 77: 7,14,…,987, 14, \dots, 98, which is 1414 numbers. Remainder 22: 2,9,…,1002, 9, \dots, 100, which is 100−27+1=15\frac{100-2}{7}+1 = 15 numbers. Total: 14+15=2914+15 = 29.

The answer is (D) 29\boxed{\textbf{(D)}\ 29}.

Pitfalls

  • Counting round-robin games as n(n−1)n(n-1) instead of (n2)\binom{n}{2}. That counts each game twice: 88 teams play 2828 games, not 5656.
  • Stopping a simulation before the cycle has repeated in full, then following a false pattern.
  • Marking a position W because one move leads to a W position. A winning move must lead to an L position.
  • Off-by-one mistakes between "after kk steps" and "at step kk." If the start is step 11, then after 55 steps you are at step 66.

Traps that recur

  • Treating the game as ordinary Nim (using wall sizes 6, 2, 1 directly) even though removing a middle brick splits a wall into two. (2021 AMC 10B #24)
  • Treating the 18 games as freely arrangeable (choosing which 3 go in each round) without enforcing that every player plays exactly once per round. (2013 AMC 10A #24)
  • Reporting the number of transitive triples, 945 (choice C), or the total number of triples, 1330 (choice E), instead of their difference. (2016 AMC 10B #22)
  • Trying to simulate all the exchanges by hand, or assuming the process ends with 0 tokens of each color. (2013 AMC 10B #17)

Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic note, tags and key insights are original to this site.