# For numbers divisible by three, why is the sum of their digits able to be divided by three? [duplicate]

When you add the digits of any number that is divisible by three, that sum of those digits also appears to be divisible by three (with no remainder).

For example a number (which I randomly grab from the top of my head):

289752

whose digits sum to 33 (2+8+9+7+5+2=33)

That sum 33 is divisible by three and so is the original number 289752.

This is not the case when dividing by 2, for example 12 is divisible by two but when its digits are summed (1+2=3) you receive 3 which is NOT divisible by 2.

I have yet to be able to find a counter example for this phenomenon of division by 3.

Why does this happen?

## marked as duplicate by user223391, Daniel FischerSep 14 '15 at 17:57

• – Brick Sep 14 '15 at 17:47

Let $n=d_0+d_1\times 10+d_2\times 10^2+\cdots d_n\times10^n$ where $d_i$ are the digits. We note that $10\equiv 1 \pmod{3}$, and thus $10^k\equiv 1\pmod{3}$, so

$$n\equiv d_0+d_1+\cdots+d_n \pmod{3}$$

So $n\equiv 0 \pmod{3}$ iff $d_0+d_1+\cdots+d_n\equiv 0 \pmod{3}$

If you don't know modular stuff, wikipedia has an article I'm sure. Just google it.

We have \begin{align}N&=a_0+a_1\cdot 10^1+a_2\cdot 10^2+\cdots+a_m\cdot 10^m\\&=a_0+(9a_1+a_1)+(99a_2+a_2)+\cdots +(99\cdots 9a_m+a_m)\\&=(9a_1+99a_2+\cdots+99\cdots 99a_m)+(a_0+a_1+a_2+a_3+\cdots+a_m)\\&=(\text{a number divisible by 3})+a_0+a_1+a_2+\cdots+a_m\end{align} So, if $N$ is a multiple of $3$, then $a_1+a_2+\cdots+a_m$ is a multiple of $3$.

If we write a positve integer x in the form $x=a_0+10a_1+...+10^na_n$ for positive integers $n,a_0,...,a_n$, we see that $x \equiv a_0+a_1+...+a_n (mod 3)$, since $10^k \equiv 1 (mod 3)$.

Thus it can be seen that $x$ is divisible by 3 if, and only if, the sum of its digits is divisible by 3.