Question

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

Answer 1

Let the 3-digit number be represented as \( abc \), where \( a, b, c \) are its digits.

We are given the condition:

\[ 2 < a \times b \times c < 7 \]

So the possible values of the product are:

\[ 3,\ 4,\ 5,\ 6 \]

Case 1: Using digits 1, 1, and x such that \(1 \times 1 \times x = x\)

Valid values for \( x \) are 3, 4, 5, and 6. That gives us 4 sets:

• (1,1,3)
• (1,1,4)
• (1,1,5)
• (1,1,6)

Each set can be arranged in 3 ways, so total = \( 4 \times 3 = 12 \) numbers

Case 2: Product = 4 using (2,2,1)

Digits = (2,2,1), which can be arranged in:

\[ \frac{3!}{2!} = 3 \text{ ways} \]

Case 3: Product = 6 using (1,2,3)

Digits = (1,2,3), all distinct. Number of permutations:

\[ 3! = 6 \text{ ways} \]

Total Count:

\[ 12 + 3 + 6 = \boxed{21} \]

Final Answer:

21 valid numbers satisfy the condition \( 2 < a \times b \times c < 7 \).