<s>[INST] Why is the sum of the digits in a multiple of 9 also a multiple of 9? The sum of the digits in $9 k$ (where $k$ is an integer) is a multiple of $9$: for example 
$$9\cdot 1=9$$
$$9\cdot 7=63 \qquad  \text{and } 6+3=9\cdot 1$$
$$9\cdot 11=99 \qquad \text{and }  9+9=9\cdot 2$$
But why?
 [/INST] 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 $9$.
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