To solve this problem, let's break it down:

Step 1: Understanding the Problem

We are asked to find how many 3-digit numbers have a digit product that is more than 2 but less than 7.

Step 2: Identify the Range for the Product

The product of the digits must be:

• Greater than 2
• Less than 7

So, the possible values for the product of the digits are 3, 4, 5, or 6.

Step 3: Identify Possible Digit Combinations

For each 3-digit number $\overline{abc}$ where $a$, $b$, and $c$ are the digits:

• $a$ must be a non-zero digit (1 through 9) since it's a 3-digit number.
• $b$ and $c$ can be any digit (0 through 9).

Let's check the combinations for the possible products:

1. Product = 3:

   • Possible digit combinations:
     • $a = 3$, $b = 1$, $c = 1$
     • $a = 1$, $b = 3$, $c = 1$
     • $a = 1$, $b = 1$, $c = 3$

2. Product = 4:

   • Possible digit combinations:
     • $a = 4$, $b = 1$, $c = 1$
     • $a = 2$, $b = 2$, $c = 1$
     • $a = 1$, $b = 4$, $c = 1$
     • $a = 1$, $b = 2$, $c = 2$

3. Product = 5:

   • Possible digit combinations:
     • $a = 5$, $b = 1$, $c = 1$
     • $a = 1$, $b = 5$, $c = 1$
     • $a = 1$, $b = 1$, $c = 5$

4. Product = 6:

   • Possible digit combinations:
     • $a = 6$, $b = 1$, $c = 1$
     • $a = 3$, $b = 2$, $c = 1$
     • $a = 2$, $b = 3$, $c = 1$
     • $a = 2$, $b = 1$, $c = 3$
     • $a = 1$, $b = 6$, $c = 1$
     • $a = 1$, $b = 3$, $c = 2$
     • $a = 1$, $b = 2$, $c = 3$

Step 4: Counting Valid 3-Digit Numbers

Let's count the total number of such 3-digit numbers:

• Product = 3: There are 3 valid combinations for each digit position, leading to $3 \times 1 = 3$ numbers.
• Product = 4: There are 4 valid combinations for each digit position, leading to $4 \times 1 = 4$ numbers.
• Product = 5: There are 3 valid combinations for each digit position, leading to $3 \times 1 = 3$ numbers.
• Product = 6: There are 7 valid combinations for each digit position, leading to $7 \times 1 = 7$ numbers.

Step 5: Sum the Valid Cases

The total number of 3-digit numbers with a product of digits more than 2 but less than 7 is: $3 + 4 + 3 + 7 = 17$

Final Answer:

There are 17 three-digit numbers for which the product of their digits is more than 2 but less than 7.

Would you like further details on any of the steps?

Here are 5 related questions for you:

1. What is the total number of 3-digit numbers?
2. How would the problem change if the product were more than 10?
3. Can you find the sum of all such 3-digit numbers with a digit product more than 2 but less than 7?
4. What if the product had to be exactly 7?
5. How many 4-digit numbers have a digit product less than 10?

Tip: Always double-check the boundaries of the conditions (like "more than 2 but less than 7") to ensure accurate counting.