Proving the sum of digits is always 9 if number is divisible by 9

The other day, I was attempting to prove that if the sum of all the digits equal to $\displaystyle 9$ then the number is divisible by $\displaystyle 9$. E.g. the number $\displaystyle 72 $— the sum is equal to nine ($\displaystyle 7 + 2 = 9$) and we know for a fact that $\displaystyle 8 \cdot 9 = 72$.

Let $\displaystyle N$ be an arbitrary number with an arbitrary number of digits. In expanded form, we can write

$\displaystyle N = a \cdot 10^b + c \cdot 10^d + e \cdot 10^f + \, \ldots \, + y \cdot 10^1 + z \cdot 10^0 \, .$

Now, assuming that the sum of all the digits equals to $\displaystyle 9$, we can write

$\displaystyle 9 = a + c + e + \, \ldots \, + y + z$

But now I'm stuck. The proof is complete if we can somehow factor the number $\displaystyle N$ by breaking out a $\displaystyle 9$ but I don't really see how I'm supposed to go about to achieve that. The binomial theorem doesn't prove very fruitful (I think) when substituting $\displaystyle 10 = 9 +1$.

Re: Proving the sum of digits is always 9 if number is divisible by 9

I am gonna try modular arithmetic...

Re: Proving the sum of digits is always 9 if number is divisible by 9

Yep, modular arithmetic was the key, I managed to prove it!

Re: Proving the sum of digits is always 9 if number is divisible by 9

Though it is really the same as working "mod 10" you can show directly that 10^n divided by 9 has remainder 1 for any n. 10^n= 999...9 (with n "9"s) plus 1.

Re: Proving the sum of digits is always 9 if number is divisible by 9

Sorry bro.If you solve it then let me know.