How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?

To solve the problem of finding how many 3-digit numbers have a product of their digits that is greater than 2 but less than 7, we can break it down step by step.

### Step 1: Understanding the Constraints
A 3-digit number is represented as \( xyz \), where:
- \( x \) is the hundreds digit, and \( x \in \{1, 2, \ldots, 9\} \)
- \( y \) is the tens digit, and \( y \in \{0, 1, \ldots, 9\} \)
- \( z \) is the units digit, and \( z \in \{0, 1, \ldots, 9\} \)

We need to find all such numbers where the product \( x \times y \times z \) satisfies the condition:
\[ 2 < x \times y \times z < 7 \]

So, the product can only take the values 3, 4, 5, or 6.

### Step 2: Find Valid Digit Combinations
We now find all possible combinations of digits \( x \), \( y \), and \( z \) such that the product of these digits is 3, 4, 5, or 6.

#### For product = 3:
Possible factorizations of 3 are:
\[ 3 \times 1 \times 1 \]

This gives possible numbers:
\[ x = 3, y = 1, z = 1 \]

The numbers formed are: **311, 131, 113**

#### For product = 4:
Possible factorizations of 4 are:
\[ 4 \times 1 \times 1, \quad 2 \times 2 \times 1 \]

The numbers formed are:
- For \( x = 4, y = 1, z = 1 \): **411, 141, 114**
- For \( x = 2, y = 2, z = 1 \): **221, 212, 122**

#### For product = 5:
Possible factorization of 5 is:
\[ 5 \times 1 \times 1 \]

The numbers formed are:
\[ 511, 151, 115 \]

#### For product = 6:
Possible factorizations of 6 are:
\[ 6 \times 1 \times 1, \quad 3 \times 2 \times 1 \]

The numbers formed are:
- For \( x = 6, y = 1, z = 1 \): **611, 161, 116**
- For \( x = 3, y = 2, z = 1 \): **321, 312, 231**

### Step 3: Total Count of Numbers
Now, let's count all the valid numbers:

- From product = 3: **3 numbers** (311, 131, 113)
- From product = 4: **6 numbers** (411, 141, 114, 221, 212, 122)
- From product = 5: **3 numbers** (511, 151, 115)
- From product = 6: **6 numbers** (611, 161, 116, 321, 312, 231)

Thus, the total number of 3-digit numbers with a product of their digits between 2 and 7 is:
\[ 3 + 6 + 3 + 6 = 18 \]

### Final Answer:
There are **18** such 3-digit numbers.

Would you like more details or have any questions?

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Here are 5 related questions to expand on this concept:
1. How many 3-digit numbers have a product of their digits greater than 10 but less than 20?
2. What is the sum of all 3-digit numbers whose digits multiply to a value between 2 and 7?
3. How many 2-digit numbers have a product of their digits greater than 2 but less than 7?
4. What is the probability of randomly selecting a 3-digit number where the product of the digits is less than 7?
5. How many 4-digit numbers satisfy a similar condition where the product of the digits is greater than 10 but less than 50?

**Tip**: Factorization helps in breaking down product constraints into manageable cases, especially for digit-based problems like this one.

Explore how to find the number of 3-digit numbers whose digits multiply to a product between 2 and 7. This problem involves concepts from number theory and factorization and includes a detailed breakdown of the possible digit combinations. Suitable for Grades 6-8.

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