AMC 8 · 2023 · #21
Grade 5 countingProblem
Alina writes the numbers 1,2,…,9 on separate cards, one number per card. She wishes to divide the cards into 3 groups of 3 cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?
Pick an answer.
AMC 8 2023 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.
Toolkit + CCSS Solution
Understand
Restated: Alina has nine cards labeled $1, 2, 3, \ldots, 9$. She wants to split them into three piles of three cards each so that all three piles have the same sum. We need to count how many different ways this split can be done.
Givens: Nine cards numbered $1$ through $9$, each number used exactly once; The cards must be divided into exactly $3$ groups of $3$ cards; The sum of the numbers in each group must be the same; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Unknowns: The number of distinct ways to partition $\{1, 2, \ldots, 9\}$ into three groups of three with equal sums
Understand
Restated: Alina has nine cards labeled $1, 2, 3, \ldots, 9$. She wants to split them into three piles of three cards each so that all three piles have the same sum. We need to count how many different ways this split can be done.
Givens: Nine cards numbered $1$ through $9$, each number used exactly once; The cards must be divided into exactly $3$ groups of $3$ cards; The sum of the numbers in each group must be the same; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The question "in how many ways…" with a small, finite setup is the classic trigger for Tool #2 (Make a Systematic List). Before listing, Tool #7 (Identify Subproblems) breaks the task into three cleaner pieces: (a) figure out what each group must sum to, (b) enumerate every triple of cards that hits that sum, (c) pick a triple-of-triples that uses each card exactly once. With the target sum pinned down, Tool #3 (Eliminate Possibilities) lets us anchor on the group that must contain the largest card $9$, then read off the remaining groups by force — almost every candidate gets ruled out, which is why the final count is so small.
Execute — Answer: C
2.NBT.B.5 Step 1 - Find the total of all nine card values.
- Adding $1 + 2 + \cdots + 9$ by pairing $1{+}9, 2{+}8, 3{+}7, 4{+}6$ gives four pairs of $10$ plus the leftover $5$.
💡 Fluent addition within $100$ — a Grade 2 skill — is enough to total nine small numbers.
3.OA.A.3 Step 2 - If the three groups are equal in sum, then each one must hold one-third of the total.
- Divide $45$ by $3$ to get the required sum per group.
💡 Sharing a total equally among three groups is a Grade 3 division word problem.
4.OA.B.4 Step 3 - List every triple of distinct cards from $\{1, \ldots, 9\}$ whose sum is $15$.
- To stay systematic, fix the largest card in the triple and work top-down.
- For largest $= 9$, the other two cards must sum to $6$: pairs $(1,5)$ and $(2,4)$ — pair $(3,3)$ is forbidden (cards are distinct).
- For largest $= 8$, the pair sums to $7$: $(1,6), (2,5), (3,4)$.
- For largest $= 7$, the pair sums to $8$: $(2,6), (3,5)$ — the pair $(1,7)$ would reuse $7$, so it is out.
- Largest $\le 6$ forces the other two to average at least $4.5$, which double-counts triples we already found.
💡 Listing all the sum-decompositions of $15$ into three distinct cards is the same idea as the Grade 4 "find all factor pairs" skill, applied to addition instead of multiplication.
5.OA.B.3 Step 4 - Now pick three triples that together use each card $1\text{-}9$ exactly once.
- The card $9$ has to land somewhere, so the partition's group containing $9$ is either $\{1,5,9\}$ or $\{2,4,9\}$.
- We split into those two cases and let the leftover cards decide the rest.
💡 Splitting into a small number of forced cases — and following the rule "the largest leftover card must appear somewhere" — is the Grade 5 standard of generating cases from a rule.
4.OA.A.3 Step 5 - Case 1, $\{1,5,9\}$ is taken.
- Remaining cards are $\{2,3,4,6,7,8\}$.
- From the Step 3 list, the only sum-$15$ triple living inside this set is $\{3,4,8\}$ (we need an $8$ from the leftovers and the only $8$-triple available is $\{3,4,8\}$, since $\{1,6,8\}$ uses the already-taken $1$ and $\{2,5,8\}$ uses the already-taken $5$).
- The last three cards $\{2,6,7\}$ sum to $15$ on their own.
- So Case 1 gives exactly one partition: $\{1,5,9\},\{3,4,8\},\{2,6,7\}$.
💡 Multi-step whole-number reasoning at Grade 4 — pick a triple, subtract, check the leftover sum — finishes this case.
4.OA.A.3 Step 6 - Case 2, $\{2,4,9\}$ is taken.
- Remaining cards are $\{1,3,5,6,7,8\}$.
- Again, the $8$ must land somewhere: from the Step 3 list, the only sum-$15$ triple containing $8$ that fits inside this set is $\{1,6,8\}$ ($\{2,5,8\}$ uses the missing $2$, $\{3,4,8\}$ uses the missing $4$).
- The leftover $\{3,5,7\}$ sums to $15$.
- So Case 2 gives exactly one partition: $\{2,4,9\},\{1,6,8\},\{3,5,7\}$.
💡 Same Grade 4 multi-step subtraction-and-check move as Case 1, applied to the second forced split.
2.NBT.B.5 Step 7 - Count the partitions found.
- Case 1 yielded one valid partition, Case 2 yielded one valid partition, and no other case is possible because card $9$ has to sit in exactly one group.
- Total: $1 + 1 = 2$ ways, matching answer choice $(C)$.
💡 Adding up the case counts is just Grade 2 addition.
2.NBT.B.5 Find the total of all nine card values. Adding $1 + 2 + \cdots + 9$ by pairing $ 3.OA.A.3 If the three groups are equal in sum, then each one must hold one-third of the t 4.OA.B.4 List every triple of distinct cards from $\{1, \ldots, 9\}$ whose sum is $15$. T 5.OA.B.3 Now pick three triples that together use each card $1\text{-}9$ exactly once. Th 4.OA.A.3 Case 1, $\{1,5,9\}$ is taken. Remaining cards are $\{2,3,4,6,7,8\}$. From the St 4.OA.A.3 Case 2, $\{2,4,9\}$ is taken. Remaining cards are $\{1,3,5,6,7,8\}$. Again, the 2.NBT.B.5 Count the partitions found. Case 1 yielded one valid partition, Case 2 yielded o Review
Reasonableness: The two partitions we found really are different — $\{1,5,9\},\{3,4,8\},\{2,6,7\}$ and $\{1,6,8\},\{2,4,9\},\{3,5,7\}$ share no group in common. Each row sums to $15$, and across the two partitions every card $1$ through $9$ appears exactly once. The answer $2$ also feels right intuitively: there are only seven triples that even sum to $15$, and once we pin down where $9$ goes, the other groups are forced — so we should expect a small number, not $3$ or $4$.
Alternative: Tool #1 (Draw a Diagram) by way of the $3 \times 3$ magic square. The unique $3 \times 3$ magic square uses the cards $1\text{-}9$ with every row, column, and diagonal summing to $15$. Its rows give the partition $\{4,9,2\},\{3,5,7\},\{8,1,6\}$ (our Case 2) and its columns give $\{4,3,8\},\{9,5,1\},\{2,7,6\}$ (our Case 1). The two diagonals also sum to $15$ but they share the center $5$, so they cannot both be groups in the same partition. This gives a slick reason why the answer is exactly $2$.
CCSS standards used (min grade 5)
2.NBT.B.5Fluently add and subtract within 100 (Totaling $1+2+\cdots+9 = 45$ and combining the case counts $1+1=2$ at the end.)3.OA.A.3Solve multiplication and division word problems within 100 (Dividing the total $45$ equally among $3$ groups to get the per-group sum $15$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Listing every triple from $\{1,\ldots,9\}$ that sums to $15$ — the same "find all decompositions" habit used for factor pairs, applied to addition.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Inside each case, picking a triple, removing those cards, and checking that the leftover three cards also sum to $15$.)5.OA.B.3Generate two numerical patterns using two given rules and identify relationships (Splitting the partition search into the two forced cases driven by the rule "the group containing $9$ must be either $\{1,5,9\}$ or $\{2,4,9\}$".)
⭐ This AMC 8 problem only needs the Grade 5 habit of generating cases from a rule — once you decide where the card $9$ goes, the rest of the groups are forced!
⭐ This AMC 8 problem only needs the Grade 5 habit of generating cases from a rule — once you decide where the card $9$ goes, the rest of the groups are forced!
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