If sum of digits of a number is divisible by 3, then the number is divisible by 3. But Why? I need the proof.

Answer 1

https://math.stackexchange.com/questions/341202/how-to-prove-the-divisibility-rule-for-3

Scrol down for the answer where it is explained

Answer 2

It is true for all number systems, that leave a remainder of 1 when divided by 3.
4, 7, 10 (decimal), 13, 16 (hexadecimal),...

I would try to explain it by first counting up to 30.
3
6
9
12
...
30
Between 0 and 30 all the sums of the digits of the numbers are divisible by 3.
Then you can do the same for 30:
30
60
90
120
...
300
Between 0 and 300 the sum of all the digits is dividible by 3.
Actually it was quite bad prove.

But it works similar for 2 and 4 in number system of 5.
because 5 % 2 = 5 % 4 = 1

Answer 3

That doesnt prove anything..
Prove is considered on writing a number in form of its digits like
252 = 100*2 + 10×5 + 1*2
Hovewer these 100 , 10 and 1 need to be represented as powers of 10, 2,1,0 in order as mentioned on that link posted above.
Proof of something needs to be in general, you cannot prove any rule by "proving" or showing something on certain range of numbers

Answer 4

Yap you are right thats the point of proof, you just need general form like for any number