How many two-digit numbers with distinct digits have a product of digits equal to 18?

Introduction:
This question involves counting two-digit numbers under digit-based constraints. You are asked to find how many two-digit numbers with distinct digits have a product of digits equal to 18. It tests logical enumeration and basic number factorization.

Given Data / Assumptions:

• We consider only two-digit numbers (from 10 to 99).
• The digits must be distinct (no repetition).
• The product of the tens digit and the units digit must be 18.
• Leading zero is not allowed in a two-digit number.

Concept / Approach:
Let the two-digit number be 10a + b, where a is the tens digit and b is the units digit. We need: a * b = 18, with the constraints:

• a is from 1 to 9 (cannot be 0 for a two-digit number).
• b is from 0 to 9.
• a and b are distinct.

We first list all factor pairs of 18 and then see which pairs fit the digit requirements.

Step-by-Step Solution:
Step 1: List factor pairs of 18 (positive digits). Possible pairs (a, b) such that a * b = 18 are: (1, 18) not valid because 18 is not a digit. (2, 9), (3, 6), (6, 3), (9, 2). Pairs involving 0 or digits above 9 are invalid.
Step 2: Check digit distinctness and two-digit structure. For each valid pair:

• (2, 9): Number = 29 (distinct digits, valid).
• (9, 2): Number = 92 (distinct digits, valid).
• (3, 6): Number = 36 (distinct digits, valid).
• (6, 3): Number = 63 (distinct digits, valid).

All four numbers are two-digit, have distinct digits, and satisfy a * b = 18.
Step 3: Count the valid numbers. The valid numbers are 29, 36, 63, and 92, so there are 4 such numbers.

Verification / Alternative Check:
Compute the products: 2 * 9 = 18 (for 29), 3 * 6 = 18 (for 36), 6 * 3 = 18 (for 63), 9 * 2 = 18 (for 92). No other two-digit combinations of distinct digits produce product 18, confirming our count.

Why Other Options Are Wrong:
Answers like 6, 8, 10, or 12 would require more two-digit combinations than actually exist. They usually come from counting invalid pairs, including repeated digits, or including numbers that are not two-digit or do not satisfy the product requirement.

Common Pitfalls:
A frequent mistake is to forget that digits must be from 0 to 9, so factor pairs such as (1, 18) are invalid. Another error is overlooking the distinct digit condition, which would incorrectly include numbers like 33 if their product matched, which it does not in this case.

Final Answer:
There are 4 such two-digit numbers with distinct digits and product of digits equal to 18.