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Question 1 : How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?


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Explanatory Answer

Let the digits of the 3-digit number be p, q, & r.
2 < p × q × r < 7
Therefore, 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 splitted, and hence if one of p,q or r is 3, the remaining two should be 1.

So, the possible combinations are
1, 1, 3
1, 3, 1
3, 1, 1
1, 1, 5
1, 5, 1
5, 1, 1

4 can be splitted as 2 × 2. Therefore, the possible combinations of p, q, r are
1, 1, 4
1, 4, 1
4, 1, 1
1, 2, 2
2, 1, 2
2, 2, 1
6 can be split as 3 × 2. Therefore, the possible combinations of p, q, r are
1, 1, 6
1, 6, 1
6, 1, 1

1, 2, 3 will also yield a product of 6. We can 3! = 6 combinations of p, q, r with 1, 2, 3
1, 2, 3
1, 3, 2
2, 1, 3
2, 3, 1
3, 1, 2
3, 2, 1

Therefore, the total number of possibilities are 3 + 3 + 3 + 3 + 3 + 6 = 21


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

Hence, the answer is, "21"

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