Show that Z=E mod 9, where E is the sum of digits of r (in dec. repres. of r)

**jmgilbert** · April 8th 2011, 03:06 PM

Hi,

I have the following problem and I don't know how to go about this... I would really appreciate if you could give me a hand. The problem says:

"Let $r \in \mathbb{N}$ . Show that

$r \equiv \Sigma mod 9$ ,

where $\Sigma$ is the sum of digits of $r$ (in decimal representation of r)."

**tonio** · April 8th 2011, 03:33 PM

Originally Posted by jmgilbert

Hi,

I have the following problem and I don't know how to go about this... I would really appreciate if you could give me a hand. The problem says:

"Let $r \in \mathbb{N}$ . Show that

$r \equiv \Sigma mod 9$ ,

where $\Sigma$ is the sum of digits of $r$ (in decimal representation of r)."

If $n=A_k\times 10^k +A_{k-1}\times 10^{k-1}+\ldots+A_1\times 10+A_0$ , and

since $10^m=1\!\!\pmod 9\,,\,\,\forall\,m\in\mathnn{N}$ , we get

$n=\sum\limits^k_{i=0}A_i\times 10^i=\sum\limits^k_{i=0}A_i\!\!\pmod 9$

Tonio

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