AMC 8 · 2025 · #3

Easy mode Grade 3

Problem

Imagine a card game called Buffalo Shuffle-o. The rule is simple: take the whole deck and split it evenly so that every player gets the same number of cards.

One day, Annika sits down to play with $3$ of her friends. That makes $4$ players in total. After the deck is split evenly, each of the $4$ players is holding $15$ cards.

For the next round, $2$ more friends join in. Now there are more players, but the deck is still the same deck. The cards get split evenly again.

How many cards does each player get this time?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: In Buffalo Shuffle-o, the whole deck is dealt evenly among the players. With Annika and $3$ friends ($4$ players total), each player gets $15$ cards. If $2$ more friends join next time so there are $6$ players, how many cards will each player get?

Givens: Game 1: $4$ players (Annika + $3$ friends); Game 1: each player is dealt $15$ cards; All cards are distributed evenly each game; Game 2 adds $2$ more friends to the same group; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $11$, (E) $12$

Unknowns: Number of cards dealt to each player in Game 2

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #8 Analyze the Units

The big question ("how many cards each in Game 2?") hides two smaller questions: (1) how big is the deck? and (2) how does that deck split among the new number of players? Tool #7 (Identify Subproblems) names those two pieces so we can solve them one at a time. Tool #8 (Analyze the Units) keeps the bookkeeping honest — $\text{players} \times \tfrac{\text{cards}}{\text{player}} = \text{cards}$, and then $\tfrac{\text{cards}}{\text{players}} = \tfrac{\text{cards}}{\text{player}}$ — which guarantees we multiply when we should and divide when we should.

Execute — Answer: C

#7 Identify Subproblems 3.OA.A.3 Step 1

Subproblem 1 — find the deck size.
In Game 1 there are $1 + 3 = 4$ players, and each is dealt $15$ cards.
Because every card in the deck is handed out, the total deck size is players times cards-per-player.

$$4 \text{ players} \times 15 \tfrac{\text{cards}}{\text{player}} = 60 \text{ cards}$$

💡 "$4$ groups of $15$" is the classic Grade 3 multiplication word-problem setup.

#7 Identify Subproblems 1.OA.A.1 Step 2

Count the new number of players.
The same $4$ players come back and $2$ more friends join, so Game 2 has $4 + 2 = 6$ players.

$$4 + 2 = 6 \text{ players}$$

💡 Adding two more to a group is a basic Grade 1 add-to word problem.

#8 Analyze the Units 3.OA.A.3 Step 3

Subproblem 2 — split the same $60$-card deck evenly among the $6$ Game 2 players.
Equal sharing is a division situation, and unit-checking confirms that $\tfrac{\text{cards}}{\text{players}}$ gives the cards-per-player we want.

$$\dfrac{60 \text{ cards}}{6 \text{ players}} = 10 \tfrac{\text{cards}}{\text{player}} \;\Rightarrow\; \textbf{(C)}$$

💡 Sharing $60$ objects equally among $6$ groups is the Grade 3 division-as-equal-sharing model.

[1] #7 3.OA.A.3 Subproblem 1 — find the deck size. In Game 1 there are $1 + 3 = 4$ players, and

[2] #7 1.OA.A.1 Count the new number of players. The same $4$ players come back and $2$ more fri

[3] #8 3.OA.A.3 Subproblem 2 — split the same $60$-card deck evenly among the $6$ Game 2 players

Review

Reasonableness: More players sharing the same deck must mean fewer cards each, so the Game 2 answer should be less than $15$. It is: $10 < 15$. Also, $10$ cards per player $\times 6$ players $= 60$ cards, which matches the deck size from Game 1 — the deck didn't shrink or grow, just got resliced. Choices (D) $11$ and (E) $12$ would need a bigger deck; (A) $8$ and (B) $9$ would leave cards left over. Only (C) $10$ divides $60$ evenly into $6$ shares.

Alternative: Tool #3 (Eliminate Possibilities) is even faster on multiple choice: the deck has $4 \times 15 = 60$ cards, and Game 2 must split it evenly into $6$ piles. Test each choice by multiplying by $6$ — only $10 \times 6 = 60$ matches, so (C). Choices that don't multiply back to $60$ ($8 \times 6 = 48$, $9 \times 6 = 54$, $11 \times 6 = 66$, $12 \times 6 = 72$) are eliminated.

CCSS standards used (min grade 3)

1.OA.A.1 Solve addition and subtraction word problems within 20 (Adding the $2$ new friends to the original $4$ players to get $6$ players in Game 2.)
3.OA.A.3 Solve multiplication and division word problems within 100 (Multiplying $4 \times 15 = 60$ to find the deck size, then dividing $60 \div 6 = 10$ to share the deck evenly among the new player count.)

⭐ This AMC 8 problem only needs Grade 3 multiplication and equal-sharing division you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 3)

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