# Why is everything multiplied by nine when the digits are added the result is nine?

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

9 * 7 = 63;         // 6 + 3 = 9
9 * 35 = 315;         // 3 + 1 + 5 = 9

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$11\cdot 9=99,9+9=18, 1111*9=9999,9+9+9+9=36?$ It should be mutiple of 9? – lab bhattacharjee Oct 4 '12 at 12:10
@labbhattacharjee The poster probably meant that you continue adding digits until you get one digit. 1+8=9 and 3+6=9 – rschwieb Oct 4 '12 at 12:17
Much more usefully, if the result of this process does produce $9$, then your original number is a multiple of $9$. – Sean Eberhard Oct 4 '12 at 12:23
In all the proofs the crucial fact is that $10\equiv 1 \ \ (9)$. – Pantelis Damianou Oct 4 '12 at 13:36

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.

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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 $a_0+10a_1+10^2a_2+\cdots 10^na_n\equiv 0\pmod{9},\quad n>0$

For every $0\leq k\leq n,\;\;10^ka_k\equiv a_k\pmod{9}$ therefore:

$$a_0+10a_1+10^2a_2+\cdots 10^na_n\equiv a_0+a_1+a_2+\cdots a_n\equiv 0\pmod{9}$$

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 $0\pmod{9}$, which is $9$.

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

Then the decimal representation of $9c$ is $\left[c-1,10-c\right]$, 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

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

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Let $p(x)$ be a polynomial, $r\in\mathbb{R}$. 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 $n\in\mathbb{N}$, 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 $n$.

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