AMC 8 · 2019 · #25

Grade 6 counting

Archetype Arithmetic Expression Decomposition └ Sub-archetype Multi-Segment Path or Count Sum

combinations-basicsystematic-enumerationlinear-diophantine identify-subproblemseasier-related-problem ↑ Prerequisites: combinations-basiclinear-diophantine

📏 Medium solution 💡 2 insights

Problem

Alice has $24$ apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?

Pick an answer.

(A)

105

(B)

114

(C)

190

(D)

210

(E)

380

AMC 8 2019 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.

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Toolkit + CCSS Solution

Understand

Restated: Alice has $24$ apples and wants to share all of them between herself, Becky, and Chris. Every person must end up with at least $2$ apples. How many different ways can the $24$ apples be split among the three of them?

Givens: There are exactly $24$ apples to share; Three people receive apples: Alice, Becky, Chris; Each person must receive at least $2$ apples; Apples are identical; people are distinct (so giving $5$ to Alice and $7$ to Becky is different from $7$ to Alice and $5$ to Becky); Answer choices: (A) $105$, (B) $114$, (C) $190$, (D) $210$, (E) $380$

Unknowns: The number of ordered triples $(a, b, c)$ of whole numbers with $a + b + c = 24$ and $a, b, c \ge 2$

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #5 Look for a Pattern, #2 Make a Systematic List, #16 Change Focus / Count the Complement

Counting ordered triples that sum to $24$ directly would mean listing roughly $200$ cases — way too many to enumerate by hand. The product-friendly path is Tool #9: first peel off the "each person gets at least $2$" rule by handing out $2$ apples to each person up front (a Tool #16-style change of focus), leaving an easier subproblem — distribute the remaining $18$ apples freely. Then attack that easier problem with Tool #9 again: replace $18$ with tiny totals ($N = 0, 1, 2, 3, \ldots$), use Tool #2 to list every triple for each small $N$, and use Tool #5 to spot the pattern in how the count grows. The pattern $1, 3, 6, 10, 15, \ldots$ — the triangular numbers — generalizes cleanly to $N = 18$.

Execute — Answer: C

#16 Change Focus / Count the Complement 4.OA.A.3 Step 1

Hand out the minimum first to make the constraint disappear.
Give $2$ apples to each of the three people up front.
That uses $2 + 2 + 2 = 6$ apples, leaving $24 - 6 = 18$ apples to share freely (each person can now receive $0$ or more extra apples).

$$24 - 2 \times 3 = 24 - 6 = 18 \text{ apples to share with no minimum}$$

💡 Pre-paying the minimum turns the hard "$\ge 2$" constraint into the simpler "$\ge 0$" problem — a Grade 4 multi-step word-problem move.

#9 Solve an Easier Related Problem 4.OA.C.5 Step 2

The easier subproblem is now: count ordered triples $(a, b, c)$ of whole numbers with $a + b + c = 18$ and $a, b, c \ge 0$.
Even $18$ feels big, so use Tool #9 — try tiny totals $N = 0, 1, 2, 3, 4$ first and list every triple for each.

Tiny cases: $N=0 \to (0,0,0)$; $N=1 \to (1,0,0),(0,1,0),(0,0,1)$; $N=2 \to (2,0,0),(0,2,0),(0,0,2),(1,1,0),(1,0,1),(0,1,1)$.

💡 Replacing $18$ by a small number we can fully list is the heart of Tool #9 — a Grade 4 "generate a pattern" skill.

#2 Make a Systematic List 4.OA.C.5 Step 3

Systematically list every triple for the small cases so the count is reliable.
Pick the ordering: sort by $a$ from largest to smallest, then by $b$.
This stops duplicates and missed cases.

$N=0$: $1$ triple. $N=1$: $3$ triples. $N=2$: $6$ triples. $N=3$: $10$ triples. $N=4$: $15$ triples.

💡 A systematic list with a fixed ordering rule is the Grade 4 way to be sure no case is skipped or double-counted.

#5 Look for a Pattern 6.EE.A.2 Step 4

Read off the pattern in the counts: $1, 3, 6, 10, 15, \ldots$ — the triangular numbers.
The count for sum $N$ is $T_{N+1} = \dfrac{(N+1)(N+2)}{2}$, because for each value of $a$ from $0$ to $N$, there are $N - a + 1$ choices for $b$ (then $c$ is forced), and $1 + 2 + 3 + \cdots + (N+1) = \dfrac{(N+1)(N+2)}{2}$.

$$\text{count}(N) = (N+1) + N + (N-1) + \cdots + 1 = \dfrac{(N+1)(N+2)}{2}$$

💡 Writing the count as an expression in $N$ — "$(N+1)(N+2)/2$" — is the Grade 6 "write expressions with letters standing for numbers" idea.

#5 Look for a Pattern 6.EE.A.4 Step 5

Verify the formula on the small cases before trusting it.
$N=0$: $\tfrac{1 \cdot 2}{2} = 1$.
$N=1$: $\tfrac{2 \cdot 3}{2} = 3$.
$N=2$: $\tfrac{3 \cdot 4}{2} = 6$.
$N=4$: $\tfrac{5 \cdot 6}{2} = 15$.
All match the listed counts.

$$\tfrac{1 \cdot 2}{2}, \tfrac{2 \cdot 3}{2}, \tfrac{3 \cdot 4}{2}, \tfrac{4 \cdot 5}{2}, \tfrac{5 \cdot 6}{2} = 1, 3, 6, 10, 15 \;\checkmark$$

💡 Checking that an expression matches the data on several inputs is exactly the Grade 6 "two expressions are equivalent" check applied to a conjectured formula.

#9 Solve an Easier Related Problem 6.EE.A.2 Step 6

Apply the formula at $N = 18$ to count distributions of the leftover $18$ apples.
This is the answer to the original sharing question, since pre-paying $2$ apples each was a one-to-one bookkeeping move.

$$\text{count}(18) = \dfrac{(18+1)(18+2)}{2} = \dfrac{19 \times 20}{2} = \dfrac{380}{2} = 190 \;\Rightarrow\; \textbf{(C)}$$

💡 Plugging $N = 18$ into the expression generalizes the small-case work to the real problem — the Grade 6 "evaluate an expression" step.

[1] #16 4.OA.A.3 Hand out the minimum first to make the constraint disappear. Give $2$ apples to

[2] #9 4.OA.C.5 The easier subproblem is now: count ordered triples $(a, b, c)$ of whole numbers

[3] #2 4.OA.C.5 Systematically list every triple for the small cases so the count is reliable. P

[4] #5 6.EE.A.2 Read off the pattern in the counts: $1, 3, 6, 10, 15, \ldots$ — the triangular n

[5] #5 6.EE.A.4 Verify the formula on the small cases before trusting it. $N=0$: $\tfrac{1 \cdot

[6] #9 6.EE.A.2 Apply the formula at $N = 18$ to count distributions of the leftover $18$ apples

Review

Reasonableness: The answer $190$ sits comfortably between answer choices that test common errors: $210 = \binom{21}{2}$ would be the count if we had forgotten to subtract one person from the divider total, and $380 = 19 \times 20$ would be the count if we forgot to divide by $2$ (i.e., counted each pair of $(a, b)$ twice as ordered $(a,b)$ vs $(b,a)$ on the wrong axis). The triangular number $\dfrac{19 \cdot 20}{2} = 190$ is also exactly $\tfrac{1}{2}$ of $380$, which is a sanity check that we paired up correctly.

Alternative: Tool #1 (Draw a Diagram) gives the classic "stars and bars" picture: lay out $18$ stars in a row and choose $2$ of the $20$ positions (including the two ends) for divider bars; the divider placements determine $(a, b, c)$. The count is $\dbinom{20}{2} = 190$, the same answer by a different route.

CCSS standards used (min grade 6)

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Pre-paying the $\ge 2$ minimum ($2 \times 3 = 6$ apples) and subtracting from $24$ to get the easier subproblem of distributing $18$ apples with no minimum.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Generating the counts $1, 3, 6, 10, 15$ for tiny totals $N = 0, 1, 2, 3, 4$ using a systematic list of triples that sum to $N$.)
6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Writing the count as the algebraic expression $\tfrac{(N+1)(N+2)}{2}$ in terms of the sum $N$, then evaluating at $N = 18$ to get $190$.)
6.EE.A.4 Identify when two expressions are equivalent (Verifying that the conjectured formula $\tfrac{(N+1)(N+2)}{2}$ matches the directly counted values for $N = 0, 1, 2, 3, 4$ before trusting it at $N = 18$.)

⭐ This AMC 8 problem only needs Grade 6 expressions and patterns you already know — list tiny cases, spot the triangular numbers, and plug in!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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