Why is the sum of the digits in a multiple of 9 also a multiple of 9?

Solution 1:

An integer $9k$ (where $k\in\Bbb{Z}$) can be written as $$9k=n_1+10n_2+100n_3+\cdots$$ where $n_1,n_2,\cdots$ are the digits of $9k$. Example: $$9\cdot145=1305=5\cdot1+10\cdot0+100\cdot3+1000\cdot1$$ Factoring, $$9k=(n_1+n_2+n_3\cdots)+(9n_2+99n_3+999n_4+\cdots)$$ $$=(n_1+n_2+n_3\cdots)+9(n_2+11n_3+111n_4+\cdots)$$ $$=\sum{\textrm{digits of 9}}+9x$$ where $x\in\Bbb{Z}$ is a number we don't really need to know. Thus, $$\sum{\textrm{digits of 9}}=9k-9x=9(k-x)$$ Since $(k-x)\in\Bbb{Z}$ then $$9\mid \sum{\textrm{digits of 9}}$$

Note: if we didn't use base 10 numbers, then 9 wouldn't have this "magic" property. If we worked in base 14 or something, then 13 would have this special property

Solution 2:

It is because $10$ gives $1$ as remainder when dividing by $9$. This can be expressed by the general notation of 'congruence' $$10\equiv 1\pmod9$$ (that expresses that these two numbers give the same remainder when dividing by $9$).

Then, it easily follows from the property of these congruences that they behave like equality with respect to addition and multiplication, that $10^n\equiv 1^n=1\pmod9$.

Alternatively, of course $10^n$ will also give remainder $1$ modulo $9$, as the number $99\dots9$ is divisible by $9$.

Now, a number $n=\overline{abc\dots}$ is just $n=\left(((a\cdot10+b)\cdot10+c)\cdot 10+\dots\right)$, so modulo $9$ we have the following congruence: $$n\equiv \left(((a\cdot 1+b)\cdot1+c)\cdot1+\dots\right)=a+b+c+\dots,$$ i.e., $n$ gives the same remainder modulo $9$ as the sum of its digits. In particular, it gives remainder $0$ (divisible by $9$) iff the sum of its digits is divisible by $9$.