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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?

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    $\begingroup$ Try writing the product as $n_0 + n_1\times 10 + n_2 \times 10^2 + \cdots + n_m \times 10^m$ where $0\leq n_i \leq 9$ are integers. Then subtract the sum of the digits (i.e. the $n_i$s) from your starting number. You can show that this difference is divisible by nine. This shows that the number you started with is divisible by nine if and only if the sum of the digits is divisible by nine. $\endgroup$
    – James
    Apr 5, 2015 at 23:16

2 Answers 2

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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

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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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