AMC 8 · 2009 · #16

Easy mode Grade 4

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Problem

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 $1 \times 3 \times 8 = 24$ . How many $3$ -digit numbers like this are there in total?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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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 $24$.

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

#7 Identify Subproblems 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 $0$".

$a, b, c \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $\; a \cdot b \cdot c = 24$

💡 Knowing that multiplying by $0$ gives $0$ — Grade 3 multiplication — instantly shrinks the candidate digits.

#2 Make a Systematic List 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 $b \cdot c = 24 / a$.

$a = 1$: $b \cdot c = 24 \Rightarrow (3, 8), (4, 6)$\\$a = 2$: $b \cdot c = 12 \Rightarrow (2, 6), (3, 4)$\\$a = 3$: $b \cdot c = 8$, but $b \ge 3$ forces $b \cdot c \ge 9$ — none\\Triples: $\{1, 3, 8\}, \{1, 4, 6\}, \{2, 2, 6\}, \{2, 3, 4\}$

💡 Systematic listing with the rule "smallest digit first" is exactly the Grade 4 "find all factor pairs" skill applied twice.

#2 Make a Systematic List 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.

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

💡 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.

#2 Make a Systematic List 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.

$\{2, 2, 6\}$: $226, 262, 622 \Rightarrow 3$

💡 With a repeated digit, only the position of the odd-one-out (the $6$) matters — $3$ slots, $3$ numbers.

#7 Identify Subproblems 4.OA.A.3 Step 5

Add up the counts from each triple to get the total.

$$6 + 6 + 6 + 3 = 21 \;\Rightarrow\; \textbf{(D)}$$

💡 Combining sub-answers (one per case) into the final total is the close-out move of Tool #7.

[1] #7 3.OA.C.7 First rule out zero. If any digit were $0$, the product would be $0$, not $24$.

[2] #2 4.OA.B.4 Find every unordered digit triple $\{a, b, c\}$ with $a \le b \le c$ and $a \cdo

[3] #2 4.OA.A.3 For each triple with three different digits, list every $3$-digit number you can

[4] #2 4.OA.A.3 For the triple $\{2, 2, 6\}$ with a repeated digit, the systematic list is short

[5] #7 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.7 Fluently 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.4 Find 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.3 Solve 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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