AMC 8 · 2009 · #16
Easy mode Grade 4Problem
Think about 3-digit whole numbers, like 135 or 226. Each one has three digits.
For each number, multiply its three digits together. We want the ones where that product equals 24.
For example, 138 works because 1×3×8=24. How many 3-digit numbers like this are there in total?
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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