How to get the value of $a + b + c$?












2












$begingroup$


$(0 leq a < b < c) in Z$,



$a + b + c + ab + ac + bc + abc = 1622$



$a + b + c = ?$



I've assumed that $a = 0$, by that we can get rid of this part $ab + ac + abc$



Now $bc + b + c = 1622$.



But I found that was useless and got stuck.






algebra-precalculus systems-of-equations







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    Hint: $(a+1)(b+1)(c+1)$
    $endgroup$
    – vadim123
    Dec 23 '18 at 16:51
















2












$begingroup$


$(0 leq a < b < c) in Z$,



$a + b + c + ab + ac + bc + abc = 1622$



$a + b + c = ?$



I've assumed that $a = 0$, by that we can get rid of this part $ab + ac + abc$



Now $bc + b + c = 1622$.



But I found that was useless and got stuck.






algebra-precalculus systems-of-equations







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    Hint: $(a+1)(b+1)(c+1)$
    $endgroup$
    – vadim123
    Dec 23 '18 at 16:51














2












2








2


1



$begingroup$


$(0 leq a < b < c) in Z$,



$a + b + c + ab + ac + bc + abc = 1622$



$a + b + c = ?$



I've assumed that $a = 0$, by that we can get rid of this part $ab + ac + abc$



Now $bc + b + c = 1622$.



But I found that was useless and got stuck.






algebra-precalculus systems-of-equations







share|cite|improve this question









$endgroup$




$(0 leq a < b < c) in Z$,



$a + b + c + ab + ac + bc + abc = 1622$



$a + b + c = ?$



I've assumed that $a = 0$, by that we can get rid of this part $ab + ac + abc$



Now $bc + b + c = 1622$.



But I found that was useless and got stuck.






algebra-precalculus systems-of-equations




algebra-precalculus systems-of-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 23 '18 at 16:50









Mohamed MagdyMohamed Magdy

19917




19917








  • 5




    $begingroup$
    Hint: $(a+1)(b+1)(c+1)$
    $endgroup$
    – vadim123
    Dec 23 '18 at 16:51














  • 5




    $begingroup$
    Hint: $(a+1)(b+1)(c+1)$
    $endgroup$
    – vadim123
    Dec 23 '18 at 16:51








5




5




$begingroup$
Hint: $(a+1)(b+1)(c+1)$
$endgroup$
– vadim123
Dec 23 '18 at 16:51




$begingroup$
Hint: $(a+1)(b+1)(c+1)$
$endgroup$
– vadim123
Dec 23 '18 at 16:51










2 Answers
2






active

oldest

votes


















3












$begingroup$

Hint:



$$a+b+c+ab+ac+bc+abc=(1+a)(1+b)(1+c)-1$$



Therefore $(1+a)(1+b)(1+c)=?$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Note that $$1+a+b+c+ab+bc+ca+abc=(1+a)(1+b)(1+c)$$
    $$(1+a)(1+b)(1+c)=1623=1cdot3cdot541$$
    $$implies a=0, b=2, c=540.$$
    Note that $0leq a<b<c$. Those three numbers, 1, 3 and 541 are the only possible factorisation that could satisfy this criteria






    share|cite|improve this answer











    $endgroup$













      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: "69"
      };
      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: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      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
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050481%2fhow-to-get-the-value-of-a-b-c%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









      3












      $begingroup$

      Hint:



      $$a+b+c+ab+ac+bc+abc=(1+a)(1+b)(1+c)-1$$



      Therefore $(1+a)(1+b)(1+c)=?$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Hint:



        $$a+b+c+ab+ac+bc+abc=(1+a)(1+b)(1+c)-1$$



        Therefore $(1+a)(1+b)(1+c)=?$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Hint:



          $$a+b+c+ab+ac+bc+abc=(1+a)(1+b)(1+c)-1$$



          Therefore $(1+a)(1+b)(1+c)=?$






          share|cite|improve this answer









          $endgroup$



          Hint:



          $$a+b+c+ab+ac+bc+abc=(1+a)(1+b)(1+c)-1$$



          Therefore $(1+a)(1+b)(1+c)=?$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 23 '18 at 16:52









          TonyKTonyK

          43k356135




          43k356135























              0












              $begingroup$

              Note that $$1+a+b+c+ab+bc+ca+abc=(1+a)(1+b)(1+c)$$
              $$(1+a)(1+b)(1+c)=1623=1cdot3cdot541$$
              $$implies a=0, b=2, c=540.$$
              Note that $0leq a<b<c$. Those three numbers, 1, 3 and 541 are the only possible factorisation that could satisfy this criteria






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Note that $$1+a+b+c+ab+bc+ca+abc=(1+a)(1+b)(1+c)$$
                $$(1+a)(1+b)(1+c)=1623=1cdot3cdot541$$
                $$implies a=0, b=2, c=540.$$
                Note that $0leq a<b<c$. Those three numbers, 1, 3 and 541 are the only possible factorisation that could satisfy this criteria






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Note that $$1+a+b+c+ab+bc+ca+abc=(1+a)(1+b)(1+c)$$
                  $$(1+a)(1+b)(1+c)=1623=1cdot3cdot541$$
                  $$implies a=0, b=2, c=540.$$
                  Note that $0leq a<b<c$. Those three numbers, 1, 3 and 541 are the only possible factorisation that could satisfy this criteria






                  share|cite|improve this answer











                  $endgroup$



                  Note that $$1+a+b+c+ab+bc+ca+abc=(1+a)(1+b)(1+c)$$
                  $$(1+a)(1+b)(1+c)=1623=1cdot3cdot541$$
                  $$implies a=0, b=2, c=540.$$
                  Note that $0leq a<b<c$. Those three numbers, 1, 3 and 541 are the only possible factorisation that could satisfy this criteria







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 23 '18 at 17:00

























                  answered Dec 23 '18 at 16:56









                  Ankit KumarAnkit Kumar

                  1,516221




                  1,516221






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics 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%2fmath.stackexchange.com%2fquestions%2f3050481%2fhow-to-get-the-value-of-a-b-c%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

                      Index of /

                      This page is only for reference, If you need detailed information, please check here
                      Read more

                      Tribalistas

                      Image
                        Nota: Para o álbum homônimo, veja Tribalistas (álbum de 2002). Para o segundo álbum homônimo, veja Tribalistas (álbum de 2017). Para o tipo de organização tribal, veja Tribalismo. Tribalistas Informação geral Origem Salvador e Rio de Janeiro País   Brasil Gênero(s) MPB Pop Samba Período em atividade 2002–2004 2017–atualmente Gravadora(s) EMI Phonomotor Records Universal Music Afiliação(ões) Dadi Pretinho da Serrinha Margareth Menezes Integrantes Arnaldo Antunes Carlinhos Brown Marisa Monte Página oficial www2.uol.com.br/tribalistas/ ...
                      Read more

                      Listed building

                      Image
                      Listed building (literalmente: prédio listado ) é toda estrutura protegida por lei pelo complexo sistema de tombamento do Reino Unido. Um listed building é qualquer estrutura oficialmente designada como tendo especial interesse arquitetônico, histórico ou cultural. É um estatuto amplamente usado, aplicado a centenas de milhares de estruturas. Estruturas tombadas não podem ser alteradas sem permissão especial, e desfrutam de isenções fiscais e outros benefícios. Cada um dos países do Reino Unido trata tombamento de forma diferente: Índice 1 Inglaterra 2 País de Gales 3 Irlanda do Norte 4 Escócia 5 Referências Inglaterra | O Palácio de Buckingham, Londres. As estruturas tombadas são listadas no Statutory List of Buildings of Special Architectural or Historic Interest , e incluem campos, sinais de trânsito, piers e áreas geográficas. A principal legislação pertinente é o Ancient Monuments and Archaeological Areas Act 1979 , [ 1 ] embora legisla...
                      Read more