AMC 8 · 2009 · #16
Grade 4 countingProblem
How many 3-digit positive integers have digits whose product equals 24?
Pick an answer.
AMC 8 2009 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: Count every $3$-digit positive integer (from $100$ to $999$) such that when you multiply its three digits together, you get $24$.
Givens: The number has exactly $3$ digits (hundreds, tens, ones); The product of the three digits equals $24$; Each digit lives in $\{0, 1, 2, \dots, 9\}$; Answer choices: (A) $12$, (B) $15$, (C) $18$, (D) $21$, (E) $24$
Unknowns: The total count of $3$-digit positive integers whose digit-product is $24$
Understand
Restated: Count every $3$-digit positive integer (from $100$ to $999$) such that when you multiply its three digits together, you get $24$.
Givens: The number has exactly $3$ digits (hundreds, tens, ones); The product of the three digits equals $24$; Each digit lives in $\{0, 1, 2, \dots, 9\}$; Answer choices: (A) $12$, (B) $15$, (C) $18$, (D) $21$, (E) $24$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems
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 $a \cdot b \cdot c = 24$, then (b) for each triple, list how many distinct $3$-digit numbers it produces. With an ordering rule $a \le b \le c$, Tool #2 finds the triples without duplicates; with another ordering rule (smallest number first), Tool #2 lists the arrangements without missing any.
Execute — Answer: D
3.OA.C.7 Step 1 - First rule out zero.
- If any digit were $0$, the product would be $0$, not $24$.
- So every digit must be from $1$ to $9$.
- That also automatically takes care of "hundreds digit isn't $0$".
💡 Knowing that multiplying by $0$ gives $0$ — Grade 3 multiplication — instantly shrinks the candidate digits.
4.OA.B.4 Step 2 - Find every unordered digit triple $\{a, b, c\}$ with $a \le b \le c$ and $a \cdot b \cdot c = 24$.
- Order by the smallest digit $a$.
- For each $a$, list the digit pairs $(b, c)$ with $b \le c$ and $b \cdot c = 24 / a$.
💡 Systematic listing with the rule "smallest digit first" is exactly the Grade 4 "find all factor pairs" skill applied twice.
4.OA.A.3 Step 3 - For each triple with three different digits, list every $3$-digit number you can form by rearranging.
- With three distinct digits, the systematic list (smallest first) yields $6$ numbers each.
💡 When the three digits are all different, picking the hundreds digit (3 choices), then the tens (2 left), then the ones (1 left) gives $3 \times 2 \times 1 = 6$ — and the list confirms it.
4.OA.A.3 Step 4 - For the triple $\{2, 2, 6\}$ with a repeated digit, the systematic list is shorter because swapping the two $2$s doesn't make a new number.
- List the positions where the $6$ can sit.
💡 With a repeated digit, only the position of the odd-one-out (the $6$) matters — $3$ slots, $3$ numbers.
4.OA.A.3 Step 5 Add up the counts from each triple to get the total.
💡 Combining sub-answers (one per case) into the final total is the close-out move of Tool #7.
3.OA.C.7 First rule out zero. If any digit were $0$, the product would be $0$, not $24$. 4.OA.B.4 Find every unordered digit triple $\{a, b, c\}$ with $a \le b \le c$ and $a \cdo 4.OA.A.3 For each triple with three different digits, list every $3$-digit number you can 4.OA.A.3 For the triple $\{2, 2, 6\}$ with a repeated digit, the systematic list is short 4.OA.A.3 Add up the counts from each triple to get the total. Review
Reasonableness: 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 ($a \le 3$ exhausts the cases since $a = 3 \Rightarrow bc = 8$ with $b \ge 3$ is impossible), so nothing was missed. $21$ matches answer choice (D).
Alternative: Tool #3 (Eliminate Possibilities) using the answer choices: notice that three triples contribute $6$ each ($3$-distinct-digit triples always give $6$) and the $\{2, 2, 6\}$ triple contributes $3$. Any combination involving the $\{2, 2, 6\}$ case forces the total to end in $\dots + 3$, so the units digit of the answer must be $3$ or $1$. Among the choices only $21$ fits the pattern $\text{(multiple of 6)} + 3$, confirming (D).
CCSS standards used (min grade 4)
3.OA.C.7Fluently multiply and divide within 100 (Recognizing that any digit equal to $0$ would force the digit product to be $0$, so every digit must be from $1$ to $9$.)4.OA.B.4Find all factor pairs for a whole number in the range 1-100 (Listing the digit triples $\{a, b, c\}$ with $a \cdot b \cdot c = 24$ by repeatedly finding the factor pairs of $24 / a$ for $a = 1, 2, 3$.)4.OA.A.3Solve multistep word problems using the four operations (Counting the arrangements per triple ($6$ for distinct-digit triples, $3$ for $\{2, 2, 6\}$) and adding them to get $6 + 6 + 6 + 3 = 21$.)
⭐ This AMC 8 problem only needs Grade 4 "list all factor pairs" plus careful counting — no permutation formula required!
⭐ This AMC 8 problem only needs Grade 4 "list all factor pairs" plus careful counting — no permutation formula required!
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