- #1

mathmari

Gold Member

MHB

- 5,049

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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)