MHB Digit sum rule for the divisibility by 9

[mathmari]

Hey!

Let $n\in \mathbb{N}$, $2\leq m\in \mathbb{N}$ and $a\in \mathbb{Z}$.

I want to show that $a\left (m+1\right )^n \overset{(9)}{\equiv} a$.

I have done the following:
\begin{equation*}a\left (m+1\right )^n \overset{(9)}{\equiv} a\left (0+1\right )^n \overset{(9)}{\equiv} a\cdot 1^n \overset{(9)}{\equiv} a\end{equation*}

Is this correct? Or do we need more details at each step? (Wondering)

After that, using the above, I want to show that \begin{equation*}\forall a_0, a_1, \ldots ,a_k\in \mathbb{Z} : \ \sum_{i=0}^ka_i10^i\overset{(9)}{\equiv}\sum_{i=0}^ka_i\end{equation*} Considering the previous result for the case $m=9$ we get:
$a(9+1)^n\overset{(9)}{\equiv}a \Rightarrow a\cdot 10^n\overset{(9)}{\equiv}a$ for $n\in \mathbb{N}$.

Let $a_0, a_1, \ldots ,a_k\in \mathbb{Z}$.

It holds the following:
\begin{equation*}\sum_{i=0}^ka_i10^i\overset{(9)}{\equiv}\sum_{i=0}^ka_i\end{equation*}
Right? (Wondering) Then how can we get from this result the digit sum rule for the divisibility of a natural number by $9$, if we consider the case $0\leq a_0, a_1, \ldots , a_k<10$ ? Could you give me a hint? (Wondering)

 

[HallsofIvy]

Your last step is correct because every a_i is less than 10 and 10= 9+ 1. So a_i(10)= a_i(9+ 1)= a_i modulo 9. The answer to your last question follows immediately from that equation: since \sum a_i 10^i= \sum a_i, modulo 9, the left side is evenly divisible by 9 if and only if the right side is: if and only if the sum of digits is a multiple of 9.

 

[I like Serena]

Hey!

Let $n\in \mathbb{N}$, $2\leq m\in \mathbb{N}$ and $a\in \mathbb{Z}$.

I want to show that $a\left (m+1\right )^n \overset{(9)}{\equiv} a$.

I have done the following:
\begin{equation*}a\left (m+1\right )^n \overset{(9)}{\equiv} a\left (0+1\right )^n \overset{(9)}{\equiv} a\cdot 1^n \overset{(9)}{\equiv} a\end{equation*}

Is this correct? Or do we need more details at each step?

Hey mathmari!

It doesn't look correct to me. (Worried)

Suppose we pick $a=3,\,m=2,\,n=1$, then we would get $3(2+1)^1=9\overset{(9)}{\equiv} 3$, but this is not true is it?

Did you perhaps mean divisibility by $m$? (Wondering)