AMC 10 · 2004 · #7
Easy mode Grade 4Three players, A, B, and C, start a game with 15, 14, and 13 tokens. In each round, whoever has the most tokens at that moment gives one token to each of the other two players and drops one more token into a discard pile. The game stops as soon as some player has no tokens left. How many rounds does the game last?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Three players start with $15$, $14$, and $13$ tokens. Each round, whoever currently has the most tokens gives one token to each of the other two players and throws one token into a discard pile. The game stops the moment any player has no tokens left. Find how many rounds are played.
Givens: Players $A$, $B$, $C$ start with $15$, $14$, $13$ tokens.; Each round the current leader gives $1$ token to each other player and discards $1$ token.; So the leader loses $3$ tokens in a round, and the two other players each gain $1$.; The game ends as soon as some player runs out of tokens (reaches $0$).; Answer choices: (A) $36$, (B) $37$, (C) $38$, (D) $39$, (E) $40$
Unknowns: The total number of rounds played before some player first reaches $0$ tokens.
Understand
Restated: Three players start with $15$, $14$, and $13$ tokens. Each round, whoever currently has the most tokens gives one token to each of the other two players and throws one token into a discard pile. The game stops the moment any player has no tokens left. Find how many rounds are played.
Givens: Players $A$, $B$, $C$ start with $15$, $14$, $13$ tokens.; Each round the current leader gives $1$ token to each other player and discards $1$ token.; So the leader loses $3$ tokens in a round, and the two other players each gain $1$.; The game ends as soon as some player runs out of tokens (reaches $0$).; Answer choices: (A) $36$, (B) $37$, (C) $38$, (D) $39$, (E) $40$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #2 Make a Systematic List
Simulating $37$ rounds one by one would work but is slow and error-prone. Instead, play just the first few rounds carefully as a systematic list (Tool #2) and watch what happens over one full turn of leadership. The three starting totals $15,14,13$ take turns being the leader, and after three rounds every player has lost exactly $1$ token while the ordering resets. That repeating three-round cycle is the pattern (Tool #5): every $3$ rounds the whole board slides down by $1$. Once you see it, you jump straight to the near-empty state instead of listing $37$ lines, then play out only the final round by hand.
Execute — Answer: B
4.NBT.B.4 Step 1 Play the first three rounds
- Write the totals as $(A,B,C)$ and track who leads.
- Start at $(15,14,13)$.
- Round 1: $A$ leads with $15$, so $A$ gives $1$ to $B$, $1$ to $C$, and discards $1$, losing $3$ to reach $12$; $B$ and $C$ each gain $1$, giving $(12,15,14)$.
- Round 2: now $B$ leads with $15$, so $B$ drops to $12$ while $A$ and $C$ each gain $1$, giving $(13,12,15)$.
- Round 3: $C$ leads with $15$ and drops to $12$ while $A$ and $B$ each gain $1$, giving $(14,13,12)$.
💡 Each of the three starting piles takes one turn as leader, so after three rounds the situation comes back around in the same order.
4.OA.C.5 Step 2 Spot the three-round cycle
- Compare the start $(15,14,13)$ with the state after three rounds $(14,13,12)$.
- Every player has exactly $1$ fewer token, and $A$ is again the leader with the other two just below.
- So the same three-round dance repeats, and each full cycle of $3$ rounds subtracts $1$ from every pile.
- That means after $3k$ rounds the totals are $(15-k,\;14-k,\;13-k)$.
💡 Three rounds later the board is a perfect copy of the start with every number lowered by one, so the pattern just repeats down the ladder.
4.OA.A.3 Step 3 Jump to the nearly-empty state
- The smallest pile is $C$, and after $3k$ rounds it holds $13-k$.
- It reaches its last safe value of $1$ token when $13-k=1$, that is $k=12$.
- Since $12$ cycles is $3\times12=36$ rounds, after $36$ rounds the totals are $(15-12,\,14-12,\,13-12)=(3,2,1)$.
- No one is empty yet, and the game continues.
💡 Extending the pattern by $12$ cycles lands you one step from the end without listing every round.
4.OA.A.3 Step 4 Play the final round
- From $(3,2,1)$ it is round $37$.
- The leader is $A$ with $3$ tokens, so $A$ must give $1$ to $B$, $1$ to $C$, and discard $1$ — giving away all $3$ tokens and landing on $0$.
- The moment $A$ reaches $0$, a player has run out, so the game ends.
- Counting the $36$ rounds from the cycles plus this last one gives $36+1=37$ rounds, which is choice (B).
💡 A leader holding exactly $3$ tokens gives them all away in one round, so that is the round the game must stop.
4.NBT.B.4 Write the totals as $(A,B,C)$ and track who leads. Start at $(15,14,13)$. Round 4.OA.C.5 Compare the start $(15,14,13)$ with the state after three rounds $(14,13,12)$. E 4.OA.A.3 The smallest pile is $C$, and after $3k$ rounds it holds $13-k$. It reaches its 4.OA.A.3 From $(3,2,1)$ it is round $37$. The leader is $A$ with $3$ tokens, so $A$ must Review
Reasonableness: Every round removes exactly $1$ token from the game ($-3$ for the leader, $+1$ each for the other two). The players start with $15+14+13=42$ tokens total. After $37$ rounds, $37$ tokens are gone, leaving $42-37=5$ tokens — and indeed the final state $(0,3,2)$ has $0+3+2=5$ tokens. The bookkeeping matches, and $37$ is choice (B).
Alternative: Track only player $A$'s tokens (Tool #4, introduce a variable to follow). Over each three-round cycle $A$ goes $15\to12\to13\to14$, a net loss of $1$, and stays the leader at the start of each cycle. After $12$ cycles $A$ has $15-12=3$ and leads, so on the next round $A$ gives away all $3$ tokens. That round is $3\times12+1=37$, confirming (B).
CCSS standards used (min grade 4)
4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Updating each player's token count round by round (leader $-3$, others $+1$) while simulating the first cycle and the final round.)4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing that every three rounds each pile drops by $1$ and generalizing to $(15-k,14-k,13-k)$ after $3k$ rounds.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Using $3\times12=36$ to jump to the near-empty state, then adding the final round to get $37$ total rounds.)
⭐ When a process repeats in a cycle, find how much changes each cycle and multiply — then you only have to play out the last little bit by hand.
⭐ When a process repeats in a cycle, find how much changes each cycle and multiply — then you only have to play out the last little bit by hand.
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