proof that the sum of digits of natural number are divisible by 3 iff the number is

Question

Im trying to prove that every natural number is divisble by three if and only if the sum of its digits are divisible by three.

First i proved by induction that $10^n-1$ is divisible by 9 (and therefore 3) so i could use this in the next step, would this be valid...

ANy natural number can be given a decimal representation

Let $s\in N$

Let $k_0,k_1...k_n\in ${${0,...10}$}

$S=k_0 +10k_1+100k_2+...k_n10^n$

$= 9k_1 + 99k_2 + ... + k_n(10^n-1) + (k_0+k_1+...k_n)$

All terms are divisible by 3 but the sum of the digits in S. Therefore Dividing S by 3 will leave the same remainder as dividing the digits bys 3.

Is this a valid proof of the claim? Thank you for your time

Answer 1

It's fine, except that after the last $=$ sign you should have written$$9k_1+99k_2+\cdots+\overbrace{99\ldots9}^{n\text{ times}}k_n+(k_0+k_1+\cdots+k_n).$$

Answer 2

Yes your proof is correct and works for both $3$ and $9$

Please polish your proof because you have missed a parenthesis somewhere in your proof.

Answer 3

Yes, don't forget your braces in $(10^n-1)$ in the last line of your equation.

In modulo arithmetic, we have $10^n \equiv (3^2+1)\equiv 1\pmod{3}$

Hence $$\sum_{i=0}^n a_i\cdot 10^i\equiv \sum_{i=0}^n a_i\cdot 1^i\equiv \sum_{i=0}^n a_i \pmod{3}$$