I want to share my observations.
We can get a remainder on division by 3 using digit sum.
For division by 3 :
As we all know if the digit sum of a given number is 3 or 6 or 9, then the number is divisible by 3 and the remainder will be 0.
If the digit sum of a given number is less than 3, then the digit sum will be the remainder.
28 divided by 3 = 9 rem 1, digit sum of 28 is 1 and remainder is also 1.
812 divided by 3 = 270 rem 2, digit sum of 812 is 2 and remainder is also 2.
If the digit sum of a given number is greater than 3, we have to repeatedly subtract 3 from the digit sum until the answer is less than 3 and that answer will be the remainder.
25 divided by 3 = 8 rem 1, digit sum of 25 is 7 and if we subtract 3 twice from 7 we get remainder as 1.
98 divided by 3 = 32 rem 2, digit sum of 98 is 8 and if we subtract 3 twice from 8 we get remainder as 2.
This holds good for any number to get a remainder on division by 3 using digit sum.
I am checking if there are any patterns to get a remainder for division by 6 using digit sum but no luck so far.