Competition · AMC preparation · step 4 of 4

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 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?

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.

How to solve

Strategy Solve an Easier Related Problem

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, …), 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, … — the triangular numbers — generalizes cleanly to N = 18.

1STEP 1

Hand out the minimum first

Give each of the three their 2 apples up front — that uses 6, leaving 18 apples to share with no minimum.

24 - 2 × 3 = 24 - 6 = 18 apples to share with no minimum

2STEP 2

Restate as a simpler count

Easier subproblem: count whole-number triples with a + b + c = 18 and each ≥ 0 — but first try tiny totals N = 0, 1, 2, 3, 4.

Tiny cases: N=0 → (0,0,0); N=1 → (1,0,0),(0,1,0),(0,0,1); N=2 → (2,0,0),(0,2,0),(0,0,2),(1,1,0),(1,0,1),(0,1,1).

3STEP 3

List the small cases

List every triple for each small case in a fixed order so nothing is missed — the counts come out 1, 3, 6, 10, 15.

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

4STEP 4

Spot the triangular numbers

Those counts 1, 3, 6, 10, 15 are the triangular numbers: for sum N the count is $\frac{(N+1)(N+2)}{2}$ .

count(N) = (N+1) + N + (N-1) + … + 1 = (N+1)(N+2)/2

5STEP 5

Check the formula on small cases

Check the formula on the small cases: $\frac{1·2}{2}=1$ , $\frac{2·3}{2}=3$ , $\frac{3·4}{2}=6$ , $\frac{5·6}{2}=15$ — all match.

(1 · 2)/2, (2 · 3)/2, (3 · 4)/2, (4 · 5)/2, (5 · 6)/2 = 1, 3, 6, 10, 15 ✓

6STEP 6

Apply the formula at 18

Plug N = 18 into the formula: $\frac{19×20}{2}$ = 190, choice (C) — the answer to the original sharing question.

count(18) = (18+1)(18+2)/2 = (19 × 20)/2 = 380/2 = 190 → (C)

Answer

190

The answer 190 sits comfortably between answer choices that test common errors: 210 = C(21, 2) would be the count if we had forgotten to subtract one person from the divider total, and 380 = 19 × 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

\frac{19 · 20}{2}

= 190 is also exactly

\frac{1}{2}

of 380, which is a sanity check that we paired up correctly.

💡Key takeaway

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

Hand out the minimum first
Restate as a simpler count
List the small cases
Spot the triangular numbers
Check the formula on small cases
Apply the formula at 18

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