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I'm working through an elementary number theory course right now and I think I've come up with a proof here but wanted some feedback on my logic.

Question: If the sum of the digits in base 10 is divisible by 9, then the number itself is divisible by 9.

Proof: Suppose that $9|d_1+d_2+...+d_n$ then $d_1+d_2+...+d_n=0\mod9$

Now consider $d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)$ Each power of $10$ is equivalent to $1\mod9$

therefor

$d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)=(1\mod9)(d_1+d_2+...d_n)$

$9|(d_1+d_2+...d_n)$ by our assumption, thus $9|(1\mod9)(d_1+d_2+...d_n)$

Thus we have shown that if 9 divides the sum of the digits in base 10, 9 divides the number itself.

The only question I really have is whether I'm jumping the gun on my assumption concerning the powers of 10 being $1\mod 9$. I think this is fair game here but not 100% confident. Thanks.

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  • $\begingroup$ Your proof is fine. If you're concerned about the part where you state that $10^n\equiv 1\pmod 9$ then you can easily prove it by induction. Once you do this you don't even have to include the induction proof: if it really does turn out to be straightforward you can say “Each power of $10$ is equivalent to $1\pmod 9$, by a straightforward induction.” $\endgroup$ – MJD Apr 2 '15 at 18:05
  • $\begingroup$ One way to see what's going on is to imagine you're given a number $n$ and you decide to write out the number $(10^n - 1)$. The result will be just a bunch of $9$'s...and a number like that will be divisible by $9$. So $10^n$ must be equivalent to $1$ mod $9$. (For a formal proof of this fact, you could use induction on $n$, as @MJD says.) $\endgroup$ – mathmandan Apr 2 '15 at 18:15
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Yes, $\,{\rm mod}\ 9\!:\ \color{}{10\equiv 1}\,\Rightarrow\, \color{#c00}{10^n}\equiv 1^n \color{#c00}{\equiv 1},\,$ therefore

$\qquad\qquad\qquad\qquad\ \ d_n \color{#c00}{10^n} + d_1\color{#c00}{10} + d_0$
$\qquad\qquad\qquad \quad \equiv\,\ d_k +\cdots + d_1 + d_0\ \ $ by $ $ basic Congruence Rules.

More efficiently, we can observe that the decimal (radix $10)$ representation of an integer $N$ is a polynomial function $\,f(10)\,$ of the radix, with integer coefficients (digits) $\,d_i,\,$ i.e.

$$\begin{eqnarray} N\, =\, f(10) \!\!\!&&= d_n 10^n +\,\cdots+d_1 10 + d_0 \\ {\rm where}\ \ f(x) \!\!&&= d_n\, x^n\,+\,\cdots\,+d_1\, x\, + d_0\end{eqnarray}$$

Thus $\ {\rm mod}\ 9\!:\,\ \color{#c00}{10\equiv 1}\,\Rightarrow\, f(\color{#c00}{10})\equiv f(\color{#c00}{1}) = d_n+\cdots + d_1 \equiv\,$ sum of digits, which follows by applying the Polynomial Congruence Rule.

The proof works for any polynomial $\,f(x)\,$ with integer coefficients. As such, these tests for divisibility by the radix$\pm1$ (e.g. also casting out nines) may be viewed as special cases of the Polynomial Congruence Rule.

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$$\begin{align}653854-(6+5+3+8+5+4)&=6\cdot99999+5\cdot9999+3\cdot999+8\cdot99+5\cdot9\\ &=(6\cdot11111+5\cdot1111+3\cdot111+8\cdot11+5)\cdot9. \end{align}$$

A number and the sum of its digits differ by a multiple of $9$.

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Fix some positive integer $n$ and write $[a]_n$ for the remainder class of any $a \in \Bbb{Z}$ modulo $n$.

It isn't hard to prove (try it!) that $[a+b]_n = [a]_n + [b]_n$ and $[ab]_n = [a]_n[b]_n$ for every $a,b \in \Bbb{Z}$. A relation with this property is called a congruence.

In particular, this means that $[a^k]_n = [a]_n^k$ for every $a,k \in \Bbb{Z}$.

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  • $\begingroup$ Could whoever down-voted this please be so kind to tell me why they did so? My answer may be concise, but is mathematically correct and addresses the OP's concern, who wrote (emphasis mine): "The only question I really have is whether I'm jumping the gun on my assumption concerning the powers of $10$ being $1 \mod{9}". $\endgroup$ – A.P. Apr 4 '15 at 12:07

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