Question: How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?
Explanation:
Let the digits of the 3-digit number be p, q, & r.
∴ 2 < p × q × r < 7
⇒ p × q × r can take the values 3, 4, 5, or 6.
Let's start with prime numbers 3 & 5.
Since they are prime, they can't be split, and hence if one of p, q or r is 3, the remaining two should be 1.
So, the possible combinations are
If p × q × r = 3
The number can be {113, 131, 311}
If p × q × r = 5
The number can be {115, 151, 511}
If p × q × r = 4
4 can be written as 1 × 2 × 2 or 1 × 1 × 4.
Therefore, the possible combinations of p, q, r are {122, 212, 221, 114, 141, 411}
If p × q × r = 6
6 can be written as 1 × 3 × 2 or 1 × 1 × 6.
Therefore, the possible combinations of p, q, r are {123, 132, 213, 231, 312, 321, 116, 161, 611}
Therefore, the total number of possibilities are 3 + 3 + 6 + 9 = 21.
Hence, 21.