Competition · AMC preparation · step 4 of 4
AMC 8 · 2025 · #3
Grade 3 arithmeticPick an answer.
AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 — players × cards/player = cards, and then cards/players = cards/player — which guarantees we multiply when we should and divide when we should.
Find the deck size
Subproblem 1 — the deck stays fixed: Game 1's 4 players × 15 cards each means the deck holds 60 cards.
"4 groups of 15" is the classic Grade 3 multiplication word-problem setup.
3.OA.A.3Identify SubproblemsCount the new players
Game 2 keeps the same 4 players and adds 2 friends, so now 6 players share the deck.
Adding two more to a group is a basic Grade 1 add-to word problem.
1.OA.A.1Identify SubproblemsSplit 60 cards among 6
Subproblem 2 — share that fixed 60-card deck among 6 players; 60 ÷ 6 gives 10 cards each → (C).
Sharing 60 objects equally among 6 groups is the Grade 3 division-as-equal-sharing model.
In Game 2 the same 60-card deck is shared equally among 6 players, so each player's card count is 60 divided by 6.
▸ Why?
The deck holds a fixed 60 cards, and that total is the same in both games.
▸ Why?
In Game 1 every card is dealt out to 4 players with 15 each, so the whole deck is 4 equal piles of 15, which is 60.
▸ Why?
Game 2 reuses the very same cards, only re-split into new piles, and cutting the 60 cards into new piles with none added or lost keeps the total at 60.
▸ Why?
Sharing that fixed 60 equally among 6 players is division, and dividing 60 by 6 tells how many cards land in each of the 6 equal piles.
▸ Why?
If each of the 6 players holds the same number of cards, the whole deck is 6 equal groups of that number.
▸ Why?
Dividing 60 by 6 exactly undoes multiplying the per-player count by 6 to rebuild the deck, so it returns that per-player count.
This AMC 8 problem only needs Grade 3 multiplication and equal-sharing division you already know!
- Find the deck size
- Count the new players
- Split 60 cards among 6
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