Question

How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7? [This Question was asked as TITA]

• A. 22
• B. 12
• C. 21
• D. 11

Answer 1

The Correct Option is C

To solve this problem, we are counting 3-digit numbers for which the product of their digits is more than 2 but less than 7. We define a 3-digit number as having digits represented by abc, where a, b, c are digits, and a ≠ 0. We examine different combinations to achieve the required product value: 3, 4, 5, or 6.
Product = 3: Possible combinations:

• 1,1,3 (numbers: 113, 131, 311)
• 1,3,1 (same as above)
• 3,1,1 (same as above)

Total: 3 numbers.
Product = 4: Possible combinations:

• 1,1,4 (numbers: 114, 141, 411)
• 1,2,2 (numbers: 122, 212, 221)

Total: 6 numbers.
Product = 5: Possible combinations:

• 1,1,5 (numbers: 115, 151, 511)

Total: 3 numbers.
Product = 6: Possible combinations:

• 1,1,6 (numbers: 116, 161, 611)
• 1,2,3 (numbers: 123, 132, 213, 231, 312, 321)
• 2,1,3 (same as above)
• 3,1,2 (same as above)
• 2,3,1 (same as above)
• 3,2,1 (same as above)

Total: 9 numbers.
Grand total: 3 + 6 + 3 + 9 = 21.
Therefore, there are 21 3-digit numbers for which the product of their digits is more than 2 but less than 7.

Answer 2

Let the 3-digit number be represented as $abc$ (where $a$, $b$, and $c$ are digits). 
We are given that the product of the digits satisfies: 
$2 < a \times b \times c < 7$.

So, the possible values of the product are: $3$, $4$, $5$, $6$.

Let’s find all 3-digit numbers whose digits multiply to one of these values:

• Product = 3: Possible digit combinations are $(1,1,3)$. Number of permutations = $3$.
• Product = 4:
  • $(1,1,4)$ → permutations = $3$
  • $(2,2,1)$ → permutations = $3$
• Product = 5: $(1,1,5)$ → permutations = $3$
• Product = 6:
  • $(1,1,6)$ → permutations = $3$
  • $(1,2,3)$ → all digits different → permutations = $3! = 6$

Total numbers = $3 + 6 + 3 + 9 = \boxed{21}$

Correct option: (C): $\boxed{21}$