Using a Do-loop to find divisors mod 13 [closed]












1












$begingroup$


I want to check sum of divisors of i mod 13 fori = 1 to i = 20. I tried writing a Do-Print loop so it looks like



Do[Print[DivisorSigma[1, i], {i, 20} mod 13]


Any help will be greatly appreciated,






core-language number-theory







share|improve this question











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closed as off-topic by Daniel Lichtblau, m_goldberg, MarcoB, eyorble, march Feb 20 at 18:53


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


  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Daniel Lichtblau, m_goldberg, MarcoB, eyorble, march

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





















    1












    $begingroup$


    I want to check sum of divisors of i mod 13 fori = 1 to i = 20. I tried writing a Do-Print loop so it looks like



    Do[Print[DivisorSigma[1, i], {i, 20} mod 13]


    Any help will be greatly appreciated,






    core-language number-theory







    share|improve this question











    $endgroup$



    closed as off-topic by Daniel Lichtblau, m_goldberg, MarcoB, eyorble, march Feb 20 at 18:53


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


    • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Daniel Lichtblau, m_goldberg, MarcoB, eyorble, march

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



















      1












      1








      1





      $begingroup$


      I want to check sum of divisors of i mod 13 fori = 1 to i = 20. I tried writing a Do-Print loop so it looks like



      Do[Print[DivisorSigma[1, i], {i, 20} mod 13]


      Any help will be greatly appreciated,






      core-language number-theory







      share|improve this question











      $endgroup$




      I want to check sum of divisors of i mod 13 fori = 1 to i = 20. I tried writing a Do-Print loop so it looks like



      Do[Print[DivisorSigma[1, i], {i, 20} mod 13]


      Any help will be greatly appreciated,






      core-language number-theory




      core-language number-theory






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Feb 19 at 23:28









      m_goldberg

      87.7k872198




      87.7k872198










      asked Feb 19 at 22:16









      argamonargamon

      294




      294




      closed as off-topic by Daniel Lichtblau, m_goldberg, MarcoB, eyorble, march Feb 20 at 18:53


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


      • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Daniel Lichtblau, m_goldberg, MarcoB, eyorble, march

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







      closed as off-topic by Daniel Lichtblau, m_goldberg, MarcoB, eyorble, march Feb 20 at 18:53


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


      • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Daniel Lichtblau, m_goldberg, MarcoB, eyorble, march

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






















          2 Answers
          2






          active

          oldest

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          2












          $begingroup$

          Table[DivisorSigma[1, i], {i, 20} ] // Mod[#, 13] &
          
(*{1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}*)


          or



          Do[Print[Mod[DivisorSigma[1, i], 13]], {i, 20}]


          which gives a column of the same numbers as before that you can look at, but it is more difficult use in further expressions.






          share|improve this answer











          $endgroup$













          • $begingroup$
            thank you so much
            $endgroup$
            – argamon
            Feb 19 at 22:28










          • $begingroup$
            You're welcome.
            $endgroup$
            – Bill Watts
            Feb 19 at 22:29



















          2












          $begingroup$

          You can do it without iterators because the functions you need (Mod, DivisorSigma) are Listable:



          Mod[DivisorSigma[1, Range[20]], 13]



          {1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}




          Note: Listable functions are applied separately to each element in a list.






          share|improve this answer









          $endgroup$













          • $begingroup$
            that's also very useful and helpful thank you
            $endgroup$
            – argamon
            Feb 19 at 22:56


















          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Table[DivisorSigma[1, i], {i, 20} ] // Mod[#, 13] &
          
(*{1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}*)


          or



          Do[Print[Mod[DivisorSigma[1, i], 13]], {i, 20}]


          which gives a column of the same numbers as before that you can look at, but it is more difficult use in further expressions.






          share|improve this answer











          $endgroup$













          • $begingroup$
            thank you so much
            $endgroup$
            – argamon
            Feb 19 at 22:28










          • $begingroup$
            You're welcome.
            $endgroup$
            – Bill Watts
            Feb 19 at 22:29
















          2












          $begingroup$

          Table[DivisorSigma[1, i], {i, 20} ] // Mod[#, 13] &
          
(*{1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}*)


          or



          Do[Print[Mod[DivisorSigma[1, i], 13]], {i, 20}]


          which gives a column of the same numbers as before that you can look at, but it is more difficult use in further expressions.






          share|improve this answer











          $endgroup$













          • $begingroup$
            thank you so much
            $endgroup$
            – argamon
            Feb 19 at 22:28










          • $begingroup$
            You're welcome.
            $endgroup$
            – Bill Watts
            Feb 19 at 22:29














          2












          2








          2





          $begingroup$

          Table[DivisorSigma[1, i], {i, 20} ] // Mod[#, 13] &
          
(*{1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}*)


          or



          Do[Print[Mod[DivisorSigma[1, i], 13]], {i, 20}]


          which gives a column of the same numbers as before that you can look at, but it is more difficult use in further expressions.






          share|improve this answer











          $endgroup$



          Table[DivisorSigma[1, i], {i, 20} ] // Mod[#, 13] &
          
(*{1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}*)


          or



          Do[Print[Mod[DivisorSigma[1, i], 13]], {i, 20}]


          which gives a column of the same numbers as before that you can look at, but it is more difficult use in further expressions.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Feb 19 at 22:33

























          answered Feb 19 at 22:27









          Bill WattsBill Watts

          3,6111621




          3,6111621












          • $begingroup$
            thank you so much
            $endgroup$
            – argamon
            Feb 19 at 22:28










          • $begingroup$
            You're welcome.
            $endgroup$
            – Bill Watts
            Feb 19 at 22:29


















          • $begingroup$
            thank you so much
            $endgroup$
            – argamon
            Feb 19 at 22:28










          • $begingroup$
            You're welcome.
            $endgroup$
            – Bill Watts
            Feb 19 at 22:29
















          $begingroup$
          thank you so much
          $endgroup$
          – argamon
          Feb 19 at 22:28




          $begingroup$
          thank you so much
          $endgroup$
          – argamon
          Feb 19 at 22:28












          $begingroup$
          You're welcome.
          $endgroup$
          – Bill Watts
          Feb 19 at 22:29




          $begingroup$
          You're welcome.
          $endgroup$
          – Bill Watts
          Feb 19 at 22:29











          2












          $begingroup$

          You can do it without iterators because the functions you need (Mod, DivisorSigma) are Listable:



          Mod[DivisorSigma[1, Range[20]], 13]



          {1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}




          Note: Listable functions are applied separately to each element in a list.






          share|improve this answer









          $endgroup$













          • $begingroup$
            that's also very useful and helpful thank you
            $endgroup$
            – argamon
            Feb 19 at 22:56
















          2












          $begingroup$

          You can do it without iterators because the functions you need (Mod, DivisorSigma) are Listable:



          Mod[DivisorSigma[1, Range[20]], 13]



          {1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}




          Note: Listable functions are applied separately to each element in a list.






          share|improve this answer









          $endgroup$













          • $begingroup$
            that's also very useful and helpful thank you
            $endgroup$
            – argamon
            Feb 19 at 22:56














          2












          2








          2





          $begingroup$

          You can do it without iterators because the functions you need (Mod, DivisorSigma) are Listable:



          Mod[DivisorSigma[1, Range[20]], 13]



          {1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}




          Note: Listable functions are applied separately to each element in a list.






          share|improve this answer









          $endgroup$



          You can do it without iterators because the functions you need (Mod, DivisorSigma) are Listable:



          Mod[DivisorSigma[1, Range[20]], 13]



          {1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}




          Note: Listable functions are applied separately to each element in a list.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Feb 19 at 22:51









          kglrkglr

          189k10206424




          189k10206424












          • $begingroup$
            that's also very useful and helpful thank you
            $endgroup$
            – argamon
            Feb 19 at 22:56


















          • $begingroup$
            that's also very useful and helpful thank you
            $endgroup$
            – argamon
            Feb 19 at 22:56
















          $begingroup$
          that's also very useful and helpful thank you
          $endgroup$
          – argamon
          Feb 19 at 22:56




          $begingroup$
          that's also very useful and helpful thank you
          $endgroup$
          – argamon
          Feb 19 at 22:56



          -

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