AMC 8 · 2025 · #3

Grade 3 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

multi-digit-arithmeticrateratio-proportion identify-subproblemsdimensional-analysis ↑ Prerequisites: multi-digit-arithmetic

📏 Short solution 💡 2 insights

📘 View easy version →

Problem

Buffalo Shuffle-o is a card game in which all the cards are distributed evenly among all players at the start of the game. When Annika and $3$ of her friends play Buffalo Shuffle-o, each player is dealt $15$ cards. Suppose $2$ more friends join the next game. How many cards will be dealt to each player?

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)

More like this