What will be the remainder if $7^{101}$ mod $5$? [closed]

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-1












$begingroup$


$$7^{101} mod 5 = 49^{100} mod 5 = (49^2)^{50} mod 5$$



What is the next step to find solution?



Regards










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$endgroup$



closed as off-topic by Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED Dec 14 '18 at 20:44


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." – Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED

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












  • 1




    $begingroup$
    $large bmod 5!: 7equiv 2, color{#c00}{2^4equiv 1},Rightarrow, 2^{4n}equiv (color{#c00}{2^4})^nequiv color{#c00}1^nequiv 1 $
    $endgroup$
    – Bill Dubuque
    Dec 14 '18 at 16:18








  • 2




    $begingroup$
    How did you get from $7^{101}$ to $49^{100}$?
    $endgroup$
    – farruhota
    Dec 14 '18 at 16:27










  • $begingroup$
    You are wrong in saying $7^{101}equiv 49^{100}$ you need $7cdot 49^{50}$ - just take care when manipulating the powers.
    $endgroup$
    – Mark Bennet
    Dec 14 '18 at 16:27






  • 1




    $begingroup$
    Not every tag which starts with the letters "mod" is about modular arithmetic. I'm surprised you didn't add (modular-forms) in addition to (modules) and (model-theory)!
    $endgroup$
    – Alex Kruckman
    Dec 14 '18 at 18:55










  • $begingroup$
    This is also a duplicate of a mother thread where the techniques for answering questions of this type have been collected. Shame on experienced users who choose to indulge in on-site-plagiarism.
    $endgroup$
    – Jyrki Lahtonen
    Dec 15 '18 at 6:13
















-1












$begingroup$


$$7^{101} mod 5 = 49^{100} mod 5 = (49^2)^{50} mod 5$$



What is the next step to find solution?



Regards










share|cite|improve this question











$endgroup$



closed as off-topic by Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED Dec 14 '18 at 20:44


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." – Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED

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












  • 1




    $begingroup$
    $large bmod 5!: 7equiv 2, color{#c00}{2^4equiv 1},Rightarrow, 2^{4n}equiv (color{#c00}{2^4})^nequiv color{#c00}1^nequiv 1 $
    $endgroup$
    – Bill Dubuque
    Dec 14 '18 at 16:18








  • 2




    $begingroup$
    How did you get from $7^{101}$ to $49^{100}$?
    $endgroup$
    – farruhota
    Dec 14 '18 at 16:27










  • $begingroup$
    You are wrong in saying $7^{101}equiv 49^{100}$ you need $7cdot 49^{50}$ - just take care when manipulating the powers.
    $endgroup$
    – Mark Bennet
    Dec 14 '18 at 16:27






  • 1




    $begingroup$
    Not every tag which starts with the letters "mod" is about modular arithmetic. I'm surprised you didn't add (modular-forms) in addition to (modules) and (model-theory)!
    $endgroup$
    – Alex Kruckman
    Dec 14 '18 at 18:55










  • $begingroup$
    This is also a duplicate of a mother thread where the techniques for answering questions of this type have been collected. Shame on experienced users who choose to indulge in on-site-plagiarism.
    $endgroup$
    – Jyrki Lahtonen
    Dec 15 '18 at 6:13














-1












-1








-1





$begingroup$


$$7^{101} mod 5 = 49^{100} mod 5 = (49^2)^{50} mod 5$$



What is the next step to find solution?



Regards










share|cite|improve this question











$endgroup$




$$7^{101} mod 5 = 49^{100} mod 5 = (49^2)^{50} mod 5$$



What is the next step to find solution?



Regards







modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 18:54









Alex Kruckman

27.6k32658




27.6k32658










asked Dec 14 '18 at 16:16









MarcinMarcin

1




1




closed as off-topic by Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED Dec 14 '18 at 20:44


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." – Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED

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







closed as off-topic by Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED Dec 14 '18 at 20:44


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." – Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED

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








  • 1




    $begingroup$
    $large bmod 5!: 7equiv 2, color{#c00}{2^4equiv 1},Rightarrow, 2^{4n}equiv (color{#c00}{2^4})^nequiv color{#c00}1^nequiv 1 $
    $endgroup$
    – Bill Dubuque
    Dec 14 '18 at 16:18








  • 2




    $begingroup$
    How did you get from $7^{101}$ to $49^{100}$?
    $endgroup$
    – farruhota
    Dec 14 '18 at 16:27










  • $begingroup$
    You are wrong in saying $7^{101}equiv 49^{100}$ you need $7cdot 49^{50}$ - just take care when manipulating the powers.
    $endgroup$
    – Mark Bennet
    Dec 14 '18 at 16:27






  • 1




    $begingroup$
    Not every tag which starts with the letters "mod" is about modular arithmetic. I'm surprised you didn't add (modular-forms) in addition to (modules) and (model-theory)!
    $endgroup$
    – Alex Kruckman
    Dec 14 '18 at 18:55










  • $begingroup$
    This is also a duplicate of a mother thread where the techniques for answering questions of this type have been collected. Shame on experienced users who choose to indulge in on-site-plagiarism.
    $endgroup$
    – Jyrki Lahtonen
    Dec 15 '18 at 6:13














  • 1




    $begingroup$
    $large bmod 5!: 7equiv 2, color{#c00}{2^4equiv 1},Rightarrow, 2^{4n}equiv (color{#c00}{2^4})^nequiv color{#c00}1^nequiv 1 $
    $endgroup$
    – Bill Dubuque
    Dec 14 '18 at 16:18








  • 2




    $begingroup$
    How did you get from $7^{101}$ to $49^{100}$?
    $endgroup$
    – farruhota
    Dec 14 '18 at 16:27










  • $begingroup$
    You are wrong in saying $7^{101}equiv 49^{100}$ you need $7cdot 49^{50}$ - just take care when manipulating the powers.
    $endgroup$
    – Mark Bennet
    Dec 14 '18 at 16:27






  • 1




    $begingroup$
    Not every tag which starts with the letters "mod" is about modular arithmetic. I'm surprised you didn't add (modular-forms) in addition to (modules) and (model-theory)!
    $endgroup$
    – Alex Kruckman
    Dec 14 '18 at 18:55










  • $begingroup$
    This is also a duplicate of a mother thread where the techniques for answering questions of this type have been collected. Shame on experienced users who choose to indulge in on-site-plagiarism.
    $endgroup$
    – Jyrki Lahtonen
    Dec 15 '18 at 6:13








1




1




$begingroup$
$large bmod 5!: 7equiv 2, color{#c00}{2^4equiv 1},Rightarrow, 2^{4n}equiv (color{#c00}{2^4})^nequiv color{#c00}1^nequiv 1 $
$endgroup$
– Bill Dubuque
Dec 14 '18 at 16:18






$begingroup$
$large bmod 5!: 7equiv 2, color{#c00}{2^4equiv 1},Rightarrow, 2^{4n}equiv (color{#c00}{2^4})^nequiv color{#c00}1^nequiv 1 $
$endgroup$
– Bill Dubuque
Dec 14 '18 at 16:18






2




2




$begingroup$
How did you get from $7^{101}$ to $49^{100}$?
$endgroup$
– farruhota
Dec 14 '18 at 16:27




$begingroup$
How did you get from $7^{101}$ to $49^{100}$?
$endgroup$
– farruhota
Dec 14 '18 at 16:27












$begingroup$
You are wrong in saying $7^{101}equiv 49^{100}$ you need $7cdot 49^{50}$ - just take care when manipulating the powers.
$endgroup$
– Mark Bennet
Dec 14 '18 at 16:27




$begingroup$
You are wrong in saying $7^{101}equiv 49^{100}$ you need $7cdot 49^{50}$ - just take care when manipulating the powers.
$endgroup$
– Mark Bennet
Dec 14 '18 at 16:27




1




1




$begingroup$
Not every tag which starts with the letters "mod" is about modular arithmetic. I'm surprised you didn't add (modular-forms) in addition to (modules) and (model-theory)!
$endgroup$
– Alex Kruckman
Dec 14 '18 at 18:55




$begingroup$
Not every tag which starts with the letters "mod" is about modular arithmetic. I'm surprised you didn't add (modular-forms) in addition to (modules) and (model-theory)!
$endgroup$
– Alex Kruckman
Dec 14 '18 at 18:55












$begingroup$
This is also a duplicate of a mother thread where the techniques for answering questions of this type have been collected. Shame on experienced users who choose to indulge in on-site-plagiarism.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 6:13




$begingroup$
This is also a duplicate of a mother thread where the techniques for answering questions of this type have been collected. Shame on experienced users who choose to indulge in on-site-plagiarism.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 6:13










3 Answers
3






active

oldest

votes


















1












$begingroup$

There is something wrong in your first step, indeed



$$7^{101} not equiv 49^{100} mod 5 ldots $$



More simply use that



$$7^2 equiv 49 equiv -1 pmod 5$$



therefore



$$7^{101} equiv 7cdot7^{100} equiv 7cdot (7^{2})^{50} equiv ,? pmod 5$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    By Carmichael function if $(a,m)=1$, then $a^{k lambda (M)}equiv 1 (mod m)$. That $lambda (5)=4$. So $7^{4k}equiv 1 (mod 5)$.



    $7^{100}equiv 1 (mod 5)$ then $7^{101}equiv 2 (mod 5)$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      use pmod{5} to render modulo.
      $endgroup$
      – zwim
      Jan 7 at 21:23





















    -1












    $begingroup$

    What if you do not see a cute way to do this?

    There is a more systematic way to do it.

    First, $7^xpmod 5$ is periodic in $x$. Compute a few of them until you see what the pattern is. Then it should be easy to find $7^{101}pmod 5$.






    share|cite|improve this answer









    $endgroup$




















      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      There is something wrong in your first step, indeed



      $$7^{101} not equiv 49^{100} mod 5 ldots $$



      More simply use that



      $$7^2 equiv 49 equiv -1 pmod 5$$



      therefore



      $$7^{101} equiv 7cdot7^{100} equiv 7cdot (7^{2})^{50} equiv ,? pmod 5$$






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        There is something wrong in your first step, indeed



        $$7^{101} not equiv 49^{100} mod 5 ldots $$



        More simply use that



        $$7^2 equiv 49 equiv -1 pmod 5$$



        therefore



        $$7^{101} equiv 7cdot7^{100} equiv 7cdot (7^{2})^{50} equiv ,? pmod 5$$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          There is something wrong in your first step, indeed



          $$7^{101} not equiv 49^{100} mod 5 ldots $$



          More simply use that



          $$7^2 equiv 49 equiv -1 pmod 5$$



          therefore



          $$7^{101} equiv 7cdot7^{100} equiv 7cdot (7^{2})^{50} equiv ,? pmod 5$$






          share|cite|improve this answer











          $endgroup$



          There is something wrong in your first step, indeed



          $$7^{101} not equiv 49^{100} mod 5 ldots $$



          More simply use that



          $$7^2 equiv 49 equiv -1 pmod 5$$



          therefore



          $$7^{101} equiv 7cdot7^{100} equiv 7cdot (7^{2})^{50} equiv ,? pmod 5$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 16:25

























          answered Dec 14 '18 at 16:20









          gimusigimusi

          92.9k84494




          92.9k84494























              0












              $begingroup$

              By Carmichael function if $(a,m)=1$, then $a^{k lambda (M)}equiv 1 (mod m)$. That $lambda (5)=4$. So $7^{4k}equiv 1 (mod 5)$.



              $7^{100}equiv 1 (mod 5)$ then $7^{101}equiv 2 (mod 5)$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                use pmod{5} to render modulo.
                $endgroup$
                – zwim
                Jan 7 at 21:23


















              0












              $begingroup$

              By Carmichael function if $(a,m)=1$, then $a^{k lambda (M)}equiv 1 (mod m)$. That $lambda (5)=4$. So $7^{4k}equiv 1 (mod 5)$.



              $7^{100}equiv 1 (mod 5)$ then $7^{101}equiv 2 (mod 5)$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                use pmod{5} to render modulo.
                $endgroup$
                – zwim
                Jan 7 at 21:23
















              0












              0








              0





              $begingroup$

              By Carmichael function if $(a,m)=1$, then $a^{k lambda (M)}equiv 1 (mod m)$. That $lambda (5)=4$. So $7^{4k}equiv 1 (mod 5)$.



              $7^{100}equiv 1 (mod 5)$ then $7^{101}equiv 2 (mod 5)$.






              share|cite|improve this answer











              $endgroup$



              By Carmichael function if $(a,m)=1$, then $a^{k lambda (M)}equiv 1 (mod m)$. That $lambda (5)=4$. So $7^{4k}equiv 1 (mod 5)$.



              $7^{100}equiv 1 (mod 5)$ then $7^{101}equiv 2 (mod 5)$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 15 '18 at 18:14

























              answered Dec 14 '18 at 18:25









              yavaryavar

              993




              993












              • $begingroup$
                use pmod{5} to render modulo.
                $endgroup$
                – zwim
                Jan 7 at 21:23




















              • $begingroup$
                use pmod{5} to render modulo.
                $endgroup$
                – zwim
                Jan 7 at 21:23


















              $begingroup$
              use pmod{5} to render modulo.
              $endgroup$
              – zwim
              Jan 7 at 21:23






              $begingroup$
              use pmod{5} to render modulo.
              $endgroup$
              – zwim
              Jan 7 at 21:23













              -1












              $begingroup$

              What if you do not see a cute way to do this?

              There is a more systematic way to do it.

              First, $7^xpmod 5$ is periodic in $x$. Compute a few of them until you see what the pattern is. Then it should be easy to find $7^{101}pmod 5$.






              share|cite|improve this answer









              $endgroup$


















                -1












                $begingroup$

                What if you do not see a cute way to do this?

                There is a more systematic way to do it.

                First, $7^xpmod 5$ is periodic in $x$. Compute a few of them until you see what the pattern is. Then it should be easy to find $7^{101}pmod 5$.






                share|cite|improve this answer









                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  What if you do not see a cute way to do this?

                  There is a more systematic way to do it.

                  First, $7^xpmod 5$ is periodic in $x$. Compute a few of them until you see what the pattern is. Then it should be easy to find $7^{101}pmod 5$.






                  share|cite|improve this answer









                  $endgroup$



                  What if you do not see a cute way to do this?

                  There is a more systematic way to do it.

                  First, $7^xpmod 5$ is periodic in $x$. Compute a few of them until you see what the pattern is. Then it should be easy to find $7^{101}pmod 5$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 14 '18 at 19:00









                  GEdgarGEdgar

                  62.7k267171




                  62.7k267171















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