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What is the remainder of Euclidean division of L=11111...1 (2018 times) in base 7 by 9?

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closed as off-topic by Leucippus, José Carlos Santos, Shailesh, Gibbs, Lee David Chung Lin Jan 29 at 0:54

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    $\begingroup$ @saulspatz in base $7$. $L =\frac{7^{2018} - 1}6$. $\endgroup$ – fleablood Jan 28 at 20:28
  • $\begingroup$ @fleablood Oops, skipped right over that. $\endgroup$ – saulspatz Jan 28 at 20:43
  • $\begingroup$ Unfortunately, not very useful in this case, as $6$ does not have an inverse mod $9$ $\endgroup$ – Jordan Green Jan 28 at 21:48
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Note that $$ \begin{split}L &= 1+7+7^2+7^3+7^4+7^5+ \dots + 7^{2017} \\ &\equiv (1+7+4)+(1+7+4)+ 7^6+\dots + 7^{2017} \pmod{9}. \end{split} $$ How many full repetitions of the pattern do we have? What is the equivalence class of each leftover term?

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$L = 111......111_7 = \sum_{i=0}^{2017} 7^i$

By Euler's Th. $7^6 \equiv 1 \pmod 9$ and so

Direct observation we can do better: $7^3\equiv(-2)^3 \equiv -8 \equiv 1 \pmod 9$

$1 + 7 + 7^2\equiv 1 -2 + 4 = 3 \pmod 9$.

So $L = \sum_{i=0}^{2017} 7^i\equiv \sum_{i=0}^{2017} 7^{i\mod 3} \equiv \sum_{i=0}^{3*672-1+2} 7^{i\mod 3}$

$\equiv \sum_{k=1}^{672} (7^0 + 7^1 + 7^2) + 7^0 + 7^1\equiv \sum_{k=1}^{672}3 + 8 \equiv 672(3) +8\equiv 8 \pmod 9$.

The remainder is $8$

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  • $\begingroup$ How is $3$ times something plus $8$ a multiple of $3$? Also I thought $2016/3$ was $672$. $\endgroup$ – Oscar Lanzi Jan 28 at 21:35
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    $\begingroup$ Good point! But $2016/6 = 336$ and $336 * 2= 772$ as everyone can do in their heads and not deign to use calculators knows.... I made arithmetic mistakes. A lot. I'll try to fix them. $\endgroup$ – fleablood Jan 28 at 22:29
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    $\begingroup$ Likewise everyone knows $3+8 = 11 \equiv 3 \mod 9$ because $9 + 3 = 11$. Duh! I mean.... that's obvious, right? $\endgroup$ – fleablood Jan 28 at 22:35
  • $\begingroup$ +1 for the sense of humor. Always good to have a proofreader. $\endgroup$ – Oscar Lanzi Jan 28 at 23:49

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