What we see: a counting question. We need 3-digit numbers whose three digits multiply to 1, 2, 3 or 4.
Step 1. Rule out zero.
What we see: the product must be a positive integer.
What it means: no digit can be 0, because then the product would be 0. So every digit is from 1 to 9.
What we do and why: list the digit groups for each product 1, 2, 3 and 4, then count the orders.
Step 2. List the digit groups.
What we see: the product is tiny, so most digits must be 1.
What it means: these are the only groups:
Product 1: 1, 1, 1
Product 2: 1, 1, 2
Product 3: 1, 1, 3
Product 4: 1, 1, 4 and 1, 2, 2
What we do and why: we checked every way to make each product with three digits, so nothing is missing.
Step 3. Count the orders of each group.
What we see: a group with all three digits the same has 1 order. A group with two digits the same has 3 orders, because the odd digit can sit in any of the 3 places.
What it means:
1, 1, 1 gives 1 number (111).
1, 1, 2 gives 3 numbers (112, 121, 211).
1, 1, 3 gives 3 numbers.
1, 1, 4 gives 3 numbers.
1, 2, 2 gives 3 numbers (122, 212, 221).
What we do and why: add them: \(1 + 3 + 3 + 3 + 3 = 13\).
The trap: forgetting that 4 can also be made as \(1 \times 2 \times 2\). That drops 3 numbers and leaves 10, which is not even a choice, so a student who misses it often guesses A (11).
Takeaway: For a small digit product, list the digit groups first, then count orders: 1 when all digits match, 3 when two match, 6 when all differ.
Answer: B
