Why is the sum of the digits in a multiple of 9 also a multiple of 9?

Question

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?

Answer 1

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

Answer 2

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 $9$.

Answer 3

It seems like this is a natural candidate for proof by induction. Suppose $n=9k$ is a multiple of $9$. Clearly the sum of the digits of $n$ are divisible by $9$ for $k=1$. Suppose that this holds for arbitrary $k$, where $n$ is the $m$-digit number given by $$n=9k=d_m d_{m-1}\ldots d_1$$ Let $s_k = d_m+\ldots+d_1$ be divisible by 9. We will show that $s_{k+1}$, which is the sum of the digits of $n+9 = 9(k+1)$ is also divisible by $9$.

Suppose $d_1=0$. Then the sum of the digits of $n+9=9(k+1)$ is $s_{k+1}=s_k+9$, which is also divisible by 9. (The ones digit $d_1$ has changed from $0$ to $9$.)

Now suppose that $d_1 > 0$ and $d_2 < 9$. Then $9(k+1) = n+9 = n + 10 - 1$. Now in the sum of digits, the ones digit $d_1$ is decreased by 1 and the tens digit $d_2$ is increased by 1, so $s_{k+1} = s_k$, which is divisible by 9.

Now if $d_2$ happens to be $9$, let $r$ be the smallest integer greater than $2$ such that $d_r < 9$. (If all digits to the left of the ones digit are $9$, then let $r = m+1$, with $d_{m+1}=0$.) In this case, all digits $d_{r-1}$ through $d_2$ are reduced by $9$ (changed from $9$ to $0$) and $d_r$ is increased by $1$, offsetting the decrease by $1$ in the ones digit. The net result is that $s_{k+1} = s_k - (r-2)\times 9$, which is also divisible by $9$.

Answer 4

Every number can be written as follows:

$$a = 10^n \cdot a_n + 10^{n-1} \cdot a_{n-1} + ... + 100 \cdot a_2 + 10 \cdot a_1 + a_0$$

(where $a_0$ is the last digit, $a_1$ the last but one, ... and $a_n$ the first)

If you subtract the sum of the digits from that number, you get:

$10^n \cdot a_n + 10^{n-1} \cdot a_{n-1} + ... + 100 \cdot a_2 + 10 \cdot a_1 + a_0 - (a_n + a_{n-1} + ... + a_2 + a_1 + a_0)$

$=(10^n-1) \cdot a_n + (10^{n-1} - 1) \cdot a_{n-1} + ... + (100-1) \cdot a_2 + (10-1) \cdot a_1$

Now take into account that $10^î-1 = 99...9$ (the digit $9$ appears $i$ times).

So that number becomes: $999...9 \cdot a_n + 99...9 \cdot a_{n-1} + ... + 99 \cdot a_2 + 9 \cdot a_1$,

which can be written as:

$$9 \times (111...1 \cdot a_n + 11...1 \cdot a_{n-1} + ... + 11 \cdot a_2 + 1 \cdot a_1)$$

Quod erat demonstrandum :-)