Suppose we have a natural number $N$ with decimal representation $A_kA_{k-1}\ldots A_0$. How do I prove that if the $\sum\limits_{i=0}^kA_i$ is divisible by $9$ then $N$ is divisible by $9$ too?
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This involves naturals or integers. Though they are reals, usually when we say reals we imply not necessarily integers. – Ross Millikan Oct 25 '13 at 16:20
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this is naturals – Nir Asaf Oct 25 '13 at 16:37
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Related : http://math.stackexchange.com/questions/328562/divisibility-criteria-for-7-11-13-17-19 – lab bhattacharjee Oct 26 '13 at 04:53
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Hint: $\overline{A_kA_{k-1}\ldots A_1A_0} = 10^kA_k + 10^{k-1}A_{k-1} + \ldots + 10A_1 + A_0$.
$(10^kA_k + 10^{k-1}A_{k-1} + \ldots + 10A_1 + A_0) - (A_k + A_{k-1} + \ldots + A_1 + A_0) = (10^k-1)A_k + (10^{k-1}-1)A_{k-1} + \ldots + (10-1)A_1$ is divisible by $9$.
njguliyev
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It would be better to show the $k-1$ term to make the pattern more obvious. You need to put braces around the subscript and superscript to make them look right. A_{k-1} gives $A_{k-1}$ – Ross Millikan Oct 25 '13 at 16:22
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What kind of example do you need? For any $m \in \mathbb{N}$ we have $9\mid 10^m-1$. – njguliyev Oct 25 '13 at 16:41
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If you are familiar with modularity you have:
- If $x \equiv a\mod{m}$ and $y \equiv b\mod{m}$ then $x+y\equiv a+b\mod{m}$.
- If $x \equiv a\mod{m}$ and $y \equiv b\mod{m}$ then $x\cdot y\equiv a\cdot b\mod{m}$.
- If $x \equiv 1\mod{m}$ then $x\cdot b\equiv b\mod{m}$.
- If $x \equiv 1\mod{m}$ then $x^i\equiv 1\mod{m}$.
- if $x\equiv0\mod m$ then $m$ divides $x$.
So if $A_kA_{k-1}\ldots A_0$ is the decimal representation of $N$, then $$N=\sum_{i=0}^k10^iA_i.$$
First we have that $10\equiv1\mod9$, and therefore (by 4.) $10^i\equiv1\mod9$
Let's evaluate it $\operatorname{mod}9$: $$N=\sum_{i=0}^k10^iA_i \equiv\sum_{i=1}^k1\cdot A_i,\qquad\text{by applying 1. 3. and 4.}$$
I let you continue from here.
