For example: I know 93681 is divisible by 3 without doing the division computation. The sum of the digits in 93681 is 9+3+6+8+1=27. If 27 is divisible by 3, then 93681 is divisible by 3. In discrete math, there was a way to prove it. How to prove that for any number, if the sum of the digits in the number is divisible by 3, then the number is divisible 3?
To show that if a natural number is divisible by three, then the sum of its digits is divisible by three, use inductive reasoning:
(A) Base case: 3 is divisible by 3, as any natural number is divisible by itself. The sum of its digits is 3, which is, of course, divisible by 3.
(B) Assumption: Let k=d(n)d(n-1)...d(1) be a multiple of 3, where the d(i) is the ith digit from the right in k. Assume that d(n)+d(n-1)+...+d(1) is divisible by 3.
(C) Induction: Now, consider k+3. One of the following occurs:
(1) 3 is added to d(1) and no carries are involved. The sum of the digits becomes d(n)+d(n-1)+...+d(1)+3, which is divisible by 3.
(2) 3 is added to d(1) and a carry to the tens place is involved. After the carry, 10 is subtracted from d(1) and 1 is added to d(2). Now, The sum of the digits becomes:
d(n)+d(n-1)+...+d(1)+3-10+1
= d(n)+d(n-1)+...+d(1)-6,
which is divisible by 3.
(3) 3 is added to d(1) and multiple carries are involved. This occurs when a string of m-1 9s ending in the tens place exists. After the carries, 10 is subtracted from d(1) and 1 is added to d(2) (making d(2) 10), 10 is subtracted from d(2) (making d(2) zero) and 1 is added to d(3), and so on, involving m carries. Then, Now, The sum of the digits becomes
d(n)+d(n-1)+...+d(1)+3-10m+m
= d(n)+d(n-1)+...+d(1)+9m-6,
which is divisible by 3.
Hence, by induction, if a natural number is divisible by three, then the sum of its digits is divisible by three.
Now, you wanted the converse. Similar proofs can show that for base cases 1 and 2, the sum of digits for k=3*p+1 and k=3*p+2 (where p is a natural number) will not be divisible by three, giving the proof that a natural number is divisible by three if and only if the sum of its digits is divisible by three.