I was walking with my girlfriend other day, and I started telling her how Indian mathematics education is quite different and taught quite differently. One example I gave her was that they teaches you that a natural number is divisible by 3 if and only if the sum of its digits (in decimal representation) are also divisible by 3. Then I asked, “do you know why?” without knowing the answer. So both of us started thinking of ways to prove it. After a moment of thoughts, I came up with the proof idea for one direction: if the sum of digits of a natural number is divisible by 3, then the number is divisible by 3. Let’s formally prove this.

Proof that if the sum of digits of a natural number in decimal representation is divisible by 3 then the number is also divisible by 3

Let n be a natural number with the decimal representation $$a_k{a}_{k-1}\cdots a_1{a}_0$$ where k is a non-negative integer and $$\mathrm{\forall }0\le i\le k,0\le a_i\le 9$$. Suppose the sum of digits is divisible by 3, which is to say  $$(\sum _{i=0}^k{a}_i)mod3\equiv 0$$

Then $$n=\sum _{i=0}^k{a}_i\cdot {10}^i$$. Because $$10mod3\equiv 1$$, it follows that $$\mathrm{\forall }i\in {\mathbb{Z}}^{+},{10}^{i}mod3\equiv 1$$ by properties of modular arithmetic. Thus $$a_i\cdot {10}^{i}mod3\equiv (a_{i}mod3)\cdot ({10}^{i}mod3)mod3\equiv a_{i}mod3$$. Hence $$nmod3\equiv (\sum _{i=0}^k{a}_i\cdot {10}^i)mod3\equiv (\sum _{i=0}^k{a}_i)mod3$$. By the supposition, $$(\sum _{i=0}^k{a}_i)mod3\equiv 0$$, and therefore $$nmod3=0$$ as desired. Q.E.D.

The proof for the other direction, however, quite troubled me to come up with. By supposing that the number is divisible by 3 doesn’t really give you any information about each digit. But this direction can be proved easily using the technique of proof by contradiction.

Proof that if a natural number is divisible by 3, then the sum of its digits in decimal representation is also divisible by 3

Suppose not. Suppose there exists a natural number n that is divisible by 3 and the sum of its digits in decimal representation is not divisible by 3. That is to say if $$a_k{a}_{k-1}\cdots a_1{a}_0$$ is the decimal representation of n, then $$(\sum _{i=0}^k{a}_i)mod3\ne 0$$. It follows immediately that $$n\equiv \sum _{i=0}^k{a}_i\cdot {10}^i$$. But $$nmod3=(\sum _{i=0}^k{a}_i)mod3$$ using the same argument made for the previous proof (this result is derived without using the divisibility of the sum of digits in the previous proof). By supposition, $$(\sum _{i=0}^k{a}_i)mod3\ne 0$$ so $$nmod3\ne 0$$. Thus n is both divisible and not divisible by 3, which is a contradiction. Thus our supposition must be false and there is no natural number divisible by 3 and the sum of its digits in decimal representation is not divisible by 3, proving the statement. Q.E.D.