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Why is it that whenever you multiply any number by nine the digits of the result always add up to a multiple of nine? Example:

9 * 7 = 63;         // 6 + 3 = 9

9 * 35 = 315;         // 3 + 1 + 5 = 9

There is a well-known divisibility test that a number is divisble by $9$ if and only if $9$ divides the sum of its digits.

By multiplying a number with $9$, you are making it a multiple of $9$, and hence the sum of its digits must be divisible by $9$. Since you can iterate the digit adding process until you get down to a single digit number, the result must be divisible by $9$.

The only two single digit numbers divisible by $9$ are $0$ and $9$, and you're never going to get $0$ (well, unless your original number was 0 :) ).

Something similar works for $3$, but it is not as nice. If you multiply a number by 3 and then iterate the digit adding process until you have one digit, then you will wind up with $3,6$ or $9$. The reason is the same: a number is divisble by 3 if and only if the sum of its digits is. However this time when you get down to single digits, there is more than one alternative that you can arrive at.

The digit sum of a positive integer is always congruent to itself modulo $9$. Since the digit sum is always less than the original number, and your original number is a mutliple of $9$, repeating the process boils it down to $9$.

Why? Write your integer as $a_0+10a_1+10^2a_2+cdots 10^na_nequiv 0pmod9,quad n>0$

For every $0leq kleq n,;;10^ka_kequiv a_kpmod9$ therefore:

$$a_0+10a_1+10^2a_2+cdots 10^na_nequiv a_0+a_1+a_2+cdots a_nequiv 0pmod9$$

However,
$$a_0+10a_1+10^2a_2+cdots 10^na_n>a_0+a_1+a_2+cdots a_n,quad (n>0)$$ Thus, repeated descent brings us to smallest positive integer congruent to $0pmod9$, which is $9$.

Let $p(x)$ be a polynomial, $rinmathbbR$. Define $q(x) := p(x)-p(r)$. Since $q(r) = 0$, we can write $q(x) = s(x)(x-r)$ for some polynomial $s(x)$. This yields

$$ p(x)-p(r) =s(x)(x-r)\
iff p(x)/(x-r) = s(x) + p(r)/(x-r) $$

Now given any number $ninmathbbN$, there is a polynomial $p_n(x)$, such that $p_n(10) = n$. Hence

$$ n/9 = p_n(10)/(10-1) = s_n(10) + p_n(1)/9$$

Hence 9 divides $n=p(10)$ if and only if 9 divides $p_n(1)$, which is the sum of all digits of $n$.

Start with
$$9cdot n= 9sum_k=0^infty a_k10^k=9(a_010^0+a_110^1+a_210^2+cdots)=9a_010^0+9a_110^1+9a_210^2+cdots$$

Then the decimal representation of $9c$ is $left[c-1,10-cright]$, if $c>0$.

So $9a_k10^k=(a_k-1)10^k+1+(10-a_k)10^k$, if $a_k>0$. Adding up all digits of the prodcut will give

$$
sum (a_k-1)+(10-a_k)=sum a_k-1+10-a_k=sum 9.
$$