AMC 8 · 2019 · #25

Grade 6 counting

Archetype Arithmetic Expression Decomposition

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?

$\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$

Pick an answer.

(A)

105

(B)

114

(C)

190

(D)

210

(E)

380

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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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