What is the remainder of Euclidean division of L=111…1 (2018 times) in base 7 by 9 [closed]












-2












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






elementary-number-theory euclidean-algorithm







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


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, José Carlos Santos, Shailesh, Gibbs, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    @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


















-2












$begingroup$


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






elementary-number-theory euclidean-algorithm







share|cite|improve this question











$endgroup$



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


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, José Carlos Santos, Shailesh, Gibbs, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    @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
















-2












-2








-2





$begingroup$


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






elementary-number-theory euclidean-algorithm







share|cite|improve this question











$endgroup$




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






elementary-number-theory euclidean-algorithm




elementary-number-theory euclidean-algorithm






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 28 at 20:26









J. W. Tanner

4,0461320




4,0461320










asked Jan 28 at 19:10









Djillali SEDjillali SE

4




4




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


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, José Carlos Santos, Shailesh, Gibbs, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.







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


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, José Carlos Santos, Shailesh, Gibbs, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    @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
















  • 1




    $begingroup$
    @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










1




1




$begingroup$
@saulspatz in base $7$. $L =frac{7^{2018} - 1}6$.
$endgroup$
– fleablood
Jan 28 at 20:28




$begingroup$
@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




$begingroup$
@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






$begingroup$
Unfortunately, not very useful in this case, as $6$ does not have an inverse mod $9$
$endgroup$
– Jordan Green
Jan 28 at 21:48












2 Answers
2






active

oldest

votes


















1












$begingroup$

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?






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

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



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



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



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



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



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



    The remainder is $8$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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








    • 1




      $begingroup$
      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








    • 1




      $begingroup$
      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




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    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?






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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?






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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?






        share|cite|improve this answer











        $endgroup$



        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?







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 28 at 21:05

























        answered Jan 28 at 20:41









        Jordan GreenJordan Green

        1,146410




        1,146410























            1












            $begingroup$

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



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



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



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



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



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



            The remainder is $8$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              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








            • 1




              $begingroup$
              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








            • 1




              $begingroup$
              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


















            1












            $begingroup$

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



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



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



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



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



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



            The remainder is $8$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              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








            • 1




              $begingroup$
              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








            • 1




              $begingroup$
              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
















            1












            1








            1





            $begingroup$

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



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



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



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



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



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



            The remainder is $8$






            share|cite|improve this answer











            $endgroup$



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



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



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



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



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



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



            The remainder is $8$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 28 at 23:51









            Oscar Lanzi

            13.3k12136




            13.3k12136










            answered Jan 28 at 20:50









            fleabloodfleablood

            73.6k22891




            73.6k22891












            • $begingroup$
              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








            • 1




              $begingroup$
              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








            • 1




              $begingroup$
              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




















            • $begingroup$
              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








            • 1




              $begingroup$
              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








            • 1




              $begingroup$
              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


















            $begingroup$
            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






            $begingroup$
            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






            1




            1




            $begingroup$
            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






            $begingroup$
            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






            1




            1




            $begingroup$
            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




            $begingroup$
            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






            $begingroup$
            +1 for the sense of humor. Always good to have a proofreader.
            $endgroup$
            – Oscar Lanzi
            Jan 28 at 23:49





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