Using a Do-loop to find divisors mod 13












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









New contributor




argamon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$

















    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









    New contributor




    argamon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      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









      New contributor




      argamon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      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









      New contributor




      argamon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      argamon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 6 hours ago









      m_goldberg

      86.4k872196




      86.4k872196






      New contributor




      argamon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 7 hours ago









      argamonargamon

      82




      82




      New contributor




      argamon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      argamon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      argamon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          2 Answers
          2






          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
            7 hours ago










          • $begingroup$
            You're welcome.
            $endgroup$
            – Bill Watts
            7 hours ago



















          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
            6 hours ago











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "387"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });






          argamon is a new contributor. Be nice, and check out our Code of Conduct.










          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f191853%2fusing-a-do-loop-to-find-divisors-mod-13%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          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
            7 hours ago










          • $begingroup$
            You're welcome.
            $endgroup$
            – Bill Watts
            7 hours ago
















          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
            7 hours ago










          • $begingroup$
            You're welcome.
            $endgroup$
            – Bill Watts
            7 hours ago














          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 7 hours ago

























          answered 7 hours ago









          Bill WattsBill Watts

          3,4311620




          3,4311620












          • $begingroup$
            thank you so much
            $endgroup$
            – argamon
            7 hours ago










          • $begingroup$
            You're welcome.
            $endgroup$
            – Bill Watts
            7 hours ago


















          • $begingroup$
            thank you so much
            $endgroup$
            – argamon
            7 hours ago










          • $begingroup$
            You're welcome.
            $endgroup$
            – Bill Watts
            7 hours ago
















          $begingroup$
          thank you so much
          $endgroup$
          – argamon
          7 hours ago




          $begingroup$
          thank you so much
          $endgroup$
          – argamon
          7 hours ago












          $begingroup$
          You're welcome.
          $endgroup$
          – Bill Watts
          7 hours ago




          $begingroup$
          You're welcome.
          $endgroup$
          – Bill Watts
          7 hours ago











          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
            6 hours ago
















          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
            6 hours ago














          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 6 hours ago









          kglrkglr

          185k10202420




          185k10202420












          • $begingroup$
            that's also very useful and helpful thank you
            $endgroup$
            – argamon
            6 hours ago


















          • $begingroup$
            that's also very useful and helpful thank you
            $endgroup$
            – argamon
            6 hours ago
















          $begingroup$
          that's also very useful and helpful thank you
          $endgroup$
          – argamon
          6 hours ago




          $begingroup$
          that's also very useful and helpful thank you
          $endgroup$
          – argamon
          6 hours ago










          argamon is a new contributor. Be nice, and check out our Code of Conduct.










          draft saved

          draft discarded


















          argamon is a new contributor. Be nice, and check out our Code of Conduct.













          argamon is a new contributor. Be nice, and check out our Code of Conduct.












          argamon is a new contributor. Be nice, and check out our Code of Conduct.
















          Thanks for contributing an answer to Mathematica Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f191853%2fusing-a-do-loop-to-find-divisors-mod-13%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Liste der Baudenkmale in Friedland (Mecklenburg)

          Image
          In der Liste der Baudenkmale in Friedland sind alle denkmalgeschützten Bauten der Stadt Friedland (Mecklenburg-Vorpommern) und ihrer Ortsteile aufgelistet. Grundlage ist die Veröffentlichung der Denkmalliste des Landkreises Mecklenburgische Seenplatte (Auszug) vom 28. September 2016. Inhaltsverzeichnis 1 Baudenkmale nach Ortsteilen 1.1 Friedland 1.2 Bresewitz 1.3 Brohm 1.4 Cosa 1.5 Dishley 1.6 Dishley-Ausbau 1.7 Eichhorst 1.8 Glienke 1.9 Heinrichshöh 1.10 Heinrichswalde 1.11 Jatzke 1.12 Jatzke-Meierei 1.13 Liepen 1.14 Ramelow 1.15 Schwanbeck 2 Vorgesehener Denkmalbereich 3 Quelle 4 Weblinks Baudenkmale nach Ortsteilen | Friedland | ID Lage Bezeichnung Beschreibung Bild 281 An der Kleinbahn 2 Windmühle mit Umgang 282 Anklamer Straße alter Friedhof mit Portalwand 282 Anklamer Straße drei Erbbegräbnissen und Gräbern u. a. H.A. Riemann 284
          Read more

          Single-Malt-Whisky

          Image
          Fünf verschiedene schottische Single-Malt-Whiskys Als Single-Malt-Whisky oder kürzer Single Malt werden Whiskys bezeichnet, die zwei besondere Bedingungen erfüllen: Sie stammen zum ersten aus einer einzigen Brennerei, sind also kein Verschnitt aus mehreren Whiskysorten (daher Single , deutsch: einzeln). Zum zweiten wird als Getreide ausschließlich gemälzte Gerste verwendet (deswegen Malt , deutsch: Malz). Der Zusatz Single Malt wird als Prädikat verstanden, die so ausgezeichneten Whiskys sind in aller Regel besonders hochwertig. Hinzu kommt ihre besonders lange Lagerung, die Reifungszeit währt oft ein Jahrzehnt und länger. Inhaltsverzeichnis 1 Allgemeines 2 Herkunftsländer 2.1 Schottland 2.2 Irland 2.3 Japan 2.4 Kontinentaleuropa 3 Produktionsschritte 3.1 Mälzen 3.2 Brauen 3.3 Brennen 3.4 Reifung und Finish 3.5 Abfüllung 4 Altersangaben 5 Lagerung von Whiskyflaschen 6 Geschmacksklassifikation 7 Sonstig
          Read more

          Czorneboh

          Image
          Dieser Artikel erläutert den Berg Czorneboh, für die Gottheit siehe Czorneboh (Gottheit). Czorneboh Čornobóh Berggasthof mit Turm Höhe 555,7  m ü. NHN [1] Lage Freistaat Sachsen, Deutschland Gebirge Lausitzer Bergland Dominanz 15,4 km →  Kottmar Koordinaten 51° 7′ 11″  N , 14° 31′ 31″  O 51.119722222222 14.525194444444 555.7 Koordinaten: 51° 7′ 11″  N , 14° 31′ 31″  O Gestein Granodiorit Besonderheiten Aussichtsturm Czorneboh Der Czorneboh , obersorbisch Čornobóh , ist ein Berg zwischen Hochkirch und Cunewalde in der Oberlausitz. Er gehört zu der südöstlich von Bautzen gelegenen Czorneboh-Bergkette und ist mit einer Höhe von 555,7  m ü. NHN [1] der höchste Punkt dieses Ausläufers des Lausitzer Berglandes. Er leitet als Landschaftselement die wellige Granitlandschaft der Lausitzer Vorbergzone ein. Der Gipfel des Czorneboh liegt in der Gemarkung Meschwitz (Gemeinde Hochkirch), die Czornebohbaude jedoch
          Read more