How many 3 digit numbers are there for which the product of their digit is note than 2 but less than 7

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.

Discover how to count the number of 3-digit numbers whose digit product is between 3 and 6. This math problem involves understanding digit products and exploring different combinations of digits. Suitable for students in Grades 4-6.

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