AMC 8 · 2009 · #16

Grade 4 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

factorssystematic-enumerationpermutations-basicdigit-constraints systematic-enumerationcasework ↑ Prerequisites: factorsmulti-digit-arithmetic

📏 Long solution 💡 4 insights

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Problem

Count every 3-digit positive integer (from 100 to 999) such that when you multiply its three digits together, you get 24.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The question "how many" with a small finite universe (3-digit numbers) is a classic Tool #2 (Systematic List) cue. Listing all 900 three-digit numbers is too many, so Tool #7 (Identify Subproblems) splits the work into two clean pieces: (a) find every unordered digit triple {a, b, c} with a · b · c = 24, then (b) for each triple, list how many distinct 3-digit numbers it produces. With an ordering rule a ≤ b ≤ c, Tool #2 finds the triples without duplicates; with another ordering rule (smallest number first), Tool #2 lists the arrangements without missing any.

1STEP 1

Any digit of 0 makes the product 0, so every digit must be 1 to 9 — that also settles "hundreds digit ≠ 0".

a, b, c ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9}, a · b · c = 24

2STEP 2

Fix smallest digit a: a = 1 → {1,3,8},{1,4,6}; a = 2 → {2,2,6},{2,3,4}; a = 3 impossible — four triples.

a = 1: b · c = 24 → (3, 8), (4, 6) ; a = 2: b · c = 12 → (2, 6), (3, 4) ; a = 3: b · c = 8, but b ≥ 3 forces b · c ≥ 9 — none ; Triples: {1, 3, 8}, {1, 4, 6}, {2, 2, 6}, {2, 3, 4}

3STEP 3

Each all-different triple ({1,3,8}, {1,4,6}, {2,3,4}) rearranges into 6 numbers — 3 × 2 × 1 slots.

{1, 3, 8}: 138, 183, 318, 381, 813, 831 → 6 ; {1, 4, 6}: 146, 164, 416, 461, 614, 641 → 6 ; {2, 3, 4}: 234, 243, 324, 342, 423, 432 → 6

4STEP 4

The repeated-digit triple {2, 2, 6} makes only 3 numbers — just the three spots the 6 can take (226, 262, 622).

{2, 2, 6}: 226, 262, 622 → 3

5STEP 5

Add the four case counts: 6 + 6 + 6 + 3 = 21 — choice (D).

6 + 6 + 6 + 3 = 21 → (D)

Answer

There are 900 three-digit numbers; we found 21 that satisfy the digit-product condition. That's about 2.3%, which feels plausible — digit-product = 24 is a fairly specific constraint, so a small fraction of three-digit numbers should qualify. Our four triples cover every way to write 24 as a product of three single digits (a ≤ 3 exhausts the cases since a = 3 → bc = 8 with b ≥ 3 is impossible), so nothing was missed. 21 matches answer choice (D).

💡Key takeaway

This AMC 8 problem only needs Grade 4 "list all factor pairs" plus careful counting — no permutation formula required!

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