Question

How many positive integers less than 1000 have tge property that the sum of the digits of each such number is divisible by 7 and tge number itself is divisible by 3?

Answer 1

answer for this question is only thirty numbers are possible.

Answer 2

 Its said the sum of digits should be divisible by 7 as well as 3. So sum of digits should be 21. Only 1 possible case

So (9-a) + (9-b) + (9-c) =21.

We need to find whole number solutions to this

So, a+b+c = 6 , hence there will be 8c2 possible numbers, or 28 such numbers

Hope it helps