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What is wrong with the logic in this proof? I can't seem to understand why it is an invalid proof.

Proposition: Let $n \in N$ represented as $d_kd_{k-1}...d_0$. Then, $9\mid n$ if and only if $9\mid \sum_{i=0}^{k} d_i$

Proof:

$$ 9\left| \sum_{i=0}^{k} d_i10^i \right.\\ 10^i\equiv 1^i \pmod 3 \\ 10^i\equiv 1 \pmod 3 \\ 9\left| \sum_{i=0}^{k} d_i *1 \right. \pmod 3 \\ 9\left| \sum_{i=0}^{k} d_i \right. \pmod 3 $$

beepbeepboop123123
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    For a start, you wrote mod3 instead of mod9. For another, you only talked about $10$ but not $10^i$. Yes, the problem is easy... but that's not an excuse to be lazy and skip steps when being exposed to proof writing for the first time. – JMoravitz Apr 02 '19 at 00:11
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    What is the logical connection between the lines in your work? – Brian61354270 Apr 02 '19 at 00:14
  • https://math.stackexchange.com/questions/328562/divisibility-criteria-for-7-11-13-17-19 – lab bhattacharjee Apr 02 '19 at 00:32
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    See here for a rigorous proof using modular arithmetic. You have some of the steps needed but you have not logically assembled them. – Bill Dubuque Apr 02 '19 at 00:53
  • I apologize, the lack of $10^i$ was a typo, as well as the $i=1$. Is mod 3 not the correct mod here ? I did not mean to exhibit the impression of being lazy. I've looked at the articles mentioned and one question I have is if $9|n\in N$ (mod3), how is it equivalent to $9|n\in N$ without the modulo? – beepbeepboop123123 Apr 02 '19 at 02:57
  • How can you make the step from line 3 to line 4? Shouldn't it be a congruence instead of a division symbol? – David Apr 02 '19 at 07:35
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    if $,{n \equiv s}\pmod{!9},$ then $,9\mid n\iff 9\mid s,,$ i.e. $,n\equiv 0\iff s\equiv 0,, $ by $,n\equiv s\pmod{!9}$ $\ \ $ – Bill Dubuque Apr 02 '19 at 15:18

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$$n=d_kd_{k-1}. . .d_2d_1=d_1+d_2\times 10+d_3\times 10^2+ . . . d_{k-1}\times 10^{k-2}+ d_k\times 10^{k-1}$$

$10^i=(9+1)^i= 9m_i +1$; $ m_i∈ N.$

⇒ $$n=d_1+d_2+ . . .+d_{k-1}+d_k+ 9(d_1m_1+d_2m_2+d_3m_3+ . . .d_{k-1}m_{k-1}+d_km_k)$$

Now if $9|n$ then we must have:

$$9|d_1+d_2+ . . .+d_{k-1}+d_k$$

sirous
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