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.