My first approach was, to make the product equal to 12 i.e. Our numbers should consists of factors of 12 (1, 2, 3, 4, 6)
Secondly, it should be divisible by 8, so it must be a ending with, (2, 4, 6)
and also, last 3 digit number should by divisible by 8
Getting A sum less than 10 is not a big deal, since we have Ones to cover it up.
Putting all the pieces together we will get our desire numbers (By using Hit and Trial)
We need to check at most, $^3C_{1}\times^4C_{2}\times2!$ combinations of last 3 digits
i.e 36
Best Answer
Let's start by using the last rule: the product of its digits is equal to 12. This is already giving us a great hint: only divisors of 12 (1, 2, 3, 4, and 6) can be digits of our number.
Next, we can think about the second clue: the sum of its digits is less than 10. Which means that, at most, our number can have 9 digits (since 0 is not a divisor of 12).
We can try to put these two clues together to see what the digits of our number look like:
Now we can try to fit the first rule in: it must be divisible by 8. This means that our number must be even and, therefore, end in a 2, a 4, or a 6.
I'll let you fill in the rest of the puzzle-pieces.