The sum of n consecutive numbers is divisible by the greatest prime factor of n.












2












$begingroup$


I facilitated the following task with pre-service math teachers:




  1. Take the sum of any three consecutive numbers. Do you notice anything special? Write a clear conjecture. Then write a clear proof for your conjecture.


  2. Now, take the sum of any amount of consecutive numbers. Can you broaden your conjecture from problem 1? Prove your conjecture.



I left the task open because I wanted students to create a variety of conjectures and proofs for whole class discussion. For task 2, one student came up with the following conjecture: "The sum of n consecutive integers is divisible by the greatest prime factor of n". I'm curious if anyone has a proof or counterexample for this claim as I do not.






number-theory







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I facilitated the following task with pre-service math teachers:




    1. Take the sum of any three consecutive numbers. Do you notice anything special? Write a clear conjecture. Then write a clear proof for your conjecture.


    2. Now, take the sum of any amount of consecutive numbers. Can you broaden your conjecture from problem 1? Prove your conjecture.



    I left the task open because I wanted students to create a variety of conjectures and proofs for whole class discussion. For task 2, one student came up with the following conjecture: "The sum of n consecutive integers is divisible by the greatest prime factor of n". I'm curious if anyone has a proof or counterexample for this claim as I do not.






    number-theory







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I facilitated the following task with pre-service math teachers:




      1. Take the sum of any three consecutive numbers. Do you notice anything special? Write a clear conjecture. Then write a clear proof for your conjecture.


      2. Now, take the sum of any amount of consecutive numbers. Can you broaden your conjecture from problem 1? Prove your conjecture.



      I left the task open because I wanted students to create a variety of conjectures and proofs for whole class discussion. For task 2, one student came up with the following conjecture: "The sum of n consecutive integers is divisible by the greatest prime factor of n". I'm curious if anyone has a proof or counterexample for this claim as I do not.






      number-theory







      share|cite|improve this question









      $endgroup$




      I facilitated the following task with pre-service math teachers:




      1. Take the sum of any three consecutive numbers. Do you notice anything special? Write a clear conjecture. Then write a clear proof for your conjecture.


      2. Now, take the sum of any amount of consecutive numbers. Can you broaden your conjecture from problem 1? Prove your conjecture.



      I left the task open because I wanted students to create a variety of conjectures and proofs for whole class discussion. For task 2, one student came up with the following conjecture: "The sum of n consecutive integers is divisible by the greatest prime factor of n". I'm curious if anyone has a proof or counterexample for this claim as I do not.






      number-theory




      number-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      MathGuyMathGuy

      913315




      913315






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          This is an excellent conjecture. It is not quite true, as it fails for $n=2$. The sum of two consecutive numbers is odd. We can say more. The sum of $n$ consecutive numbers is divisible by $n$ if $n$ is odd and by $frac n2$ if $n$ is even. This implies the student's conjecture for $n gt 2$.



          To see this, reduce all the numbers $bmod n$. We will then have one each congruent to $0,1,2,ldots n-1 bmod n$. The sum of the numbers from $0$ to $n-1$ is $frac 12(n-1)n$, which is divisible by $n$ or $frac n2$ as required.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            If $n=2$ the statement is false. Let's look at $n>2$.



            The sum of $n$ consecutive numbers starting with $a$ is



            $$
            z=frac{n}{2}(2a+n-1)
            $$
            If $n$ is even, $n/2$ is an integer containing the largest prime factor of $n$, hence $z$ is divisible by that prime factor.



            If $n$ is odd, $2a-1+n$ is even and $(2a+n-1)/2$ is an integer. Therefore $z$ is divisible by $n$ and by all of its prime factors.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              If the first of the $n$ summands is $m+1$, then the sum is
              $$(m+1)+(m+2)+ldots+(m+n)=nm+1+2+ldots+n=nm+frac{n(n+1)}{2}. $$




              • If $n$ is odd, say $n=2k-1$, this is even a multiple of $n$, namely $ncdot(m+k)$. Then even more so, it is a multiple of e.g. the largest prime divisor of $n$.

              • If $n$ is even, say $n=2k$, then it is at least a multiple of $k$, namely $kcdot(m+n+1)$. This is still a multiple of the largest prime divisor of $n$, unless $k=1$.


              Hence the conjecture fails only for $n=2$ (and is meaningless for $n=1$).






              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%2f3112297%2fthe-sum-of-n-consecutive-numbers-is-divisible-by-the-greatest-prime-factor-of-n%23new-answer', 'question_page');
                }
                );

                Post as a guest















                Required, but never shown

























                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                4












                $begingroup$

                This is an excellent conjecture. It is not quite true, as it fails for $n=2$. The sum of two consecutive numbers is odd. We can say more. The sum of $n$ consecutive numbers is divisible by $n$ if $n$ is odd and by $frac n2$ if $n$ is even. This implies the student's conjecture for $n gt 2$.



                To see this, reduce all the numbers $bmod n$. We will then have one each congruent to $0,1,2,ldots n-1 bmod n$. The sum of the numbers from $0$ to $n-1$ is $frac 12(n-1)n$, which is divisible by $n$ or $frac n2$ as required.






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  This is an excellent conjecture. It is not quite true, as it fails for $n=2$. The sum of two consecutive numbers is odd. We can say more. The sum of $n$ consecutive numbers is divisible by $n$ if $n$ is odd and by $frac n2$ if $n$ is even. This implies the student's conjecture for $n gt 2$.



                  To see this, reduce all the numbers $bmod n$. We will then have one each congruent to $0,1,2,ldots n-1 bmod n$. The sum of the numbers from $0$ to $n-1$ is $frac 12(n-1)n$, which is divisible by $n$ or $frac n2$ as required.






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    This is an excellent conjecture. It is not quite true, as it fails for $n=2$. The sum of two consecutive numbers is odd. We can say more. The sum of $n$ consecutive numbers is divisible by $n$ if $n$ is odd and by $frac n2$ if $n$ is even. This implies the student's conjecture for $n gt 2$.



                    To see this, reduce all the numbers $bmod n$. We will then have one each congruent to $0,1,2,ldots n-1 bmod n$. The sum of the numbers from $0$ to $n-1$ is $frac 12(n-1)n$, which is divisible by $n$ or $frac n2$ as required.






                    share|cite|improve this answer









                    $endgroup$



                    This is an excellent conjecture. It is not quite true, as it fails for $n=2$. The sum of two consecutive numbers is odd. We can say more. The sum of $n$ consecutive numbers is divisible by $n$ if $n$ is odd and by $frac n2$ if $n$ is even. This implies the student's conjecture for $n gt 2$.



                    To see this, reduce all the numbers $bmod n$. We will then have one each congruent to $0,1,2,ldots n-1 bmod n$. The sum of the numbers from $0$ to $n-1$ is $frac 12(n-1)n$, which is divisible by $n$ or $frac n2$ as required.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 56 mins ago









                    Ross MillikanRoss Millikan

                    296k23198371




                    296k23198371























                        2












                        $begingroup$

                        If $n=2$ the statement is false. Let's look at $n>2$.



                        The sum of $n$ consecutive numbers starting with $a$ is



                        $$
                        z=frac{n}{2}(2a+n-1)
                        $$
                        If $n$ is even, $n/2$ is an integer containing the largest prime factor of $n$, hence $z$ is divisible by that prime factor.



                        If $n$ is odd, $2a-1+n$ is even and $(2a+n-1)/2$ is an integer. Therefore $z$ is divisible by $n$ and by all of its prime factors.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          If $n=2$ the statement is false. Let's look at $n>2$.



                          The sum of $n$ consecutive numbers starting with $a$ is



                          $$
                          z=frac{n}{2}(2a+n-1)
                          $$
                          If $n$ is even, $n/2$ is an integer containing the largest prime factor of $n$, hence $z$ is divisible by that prime factor.



                          If $n$ is odd, $2a-1+n$ is even and $(2a+n-1)/2$ is an integer. Therefore $z$ is divisible by $n$ and by all of its prime factors.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            If $n=2$ the statement is false. Let's look at $n>2$.



                            The sum of $n$ consecutive numbers starting with $a$ is



                            $$
                            z=frac{n}{2}(2a+n-1)
                            $$
                            If $n$ is even, $n/2$ is an integer containing the largest prime factor of $n$, hence $z$ is divisible by that prime factor.



                            If $n$ is odd, $2a-1+n$ is even and $(2a+n-1)/2$ is an integer. Therefore $z$ is divisible by $n$ and by all of its prime factors.






                            share|cite|improve this answer









                            $endgroup$



                            If $n=2$ the statement is false. Let's look at $n>2$.



                            The sum of $n$ consecutive numbers starting with $a$ is



                            $$
                            z=frac{n}{2}(2a+n-1)
                            $$
                            If $n$ is even, $n/2$ is an integer containing the largest prime factor of $n$, hence $z$ is divisible by that prime factor.



                            If $n$ is odd, $2a-1+n$ is even and $(2a+n-1)/2$ is an integer. Therefore $z$ is divisible by $n$ and by all of its prime factors.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 47 mins ago









                            GReyesGReyes

                            1,23915




                            1,23915























                                1












                                $begingroup$

                                If the first of the $n$ summands is $m+1$, then the sum is
                                $$(m+1)+(m+2)+ldots+(m+n)=nm+1+2+ldots+n=nm+frac{n(n+1)}{2}. $$




                                • If $n$ is odd, say $n=2k-1$, this is even a multiple of $n$, namely $ncdot(m+k)$. Then even more so, it is a multiple of e.g. the largest prime divisor of $n$.

                                • If $n$ is even, say $n=2k$, then it is at least a multiple of $k$, namely $kcdot(m+n+1)$. This is still a multiple of the largest prime divisor of $n$, unless $k=1$.


                                Hence the conjecture fails only for $n=2$ (and is meaningless for $n=1$).






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  If the first of the $n$ summands is $m+1$, then the sum is
                                  $$(m+1)+(m+2)+ldots+(m+n)=nm+1+2+ldots+n=nm+frac{n(n+1)}{2}. $$




                                  • If $n$ is odd, say $n=2k-1$, this is even a multiple of $n$, namely $ncdot(m+k)$. Then even more so, it is a multiple of e.g. the largest prime divisor of $n$.

                                  • If $n$ is even, say $n=2k$, then it is at least a multiple of $k$, namely $kcdot(m+n+1)$. This is still a multiple of the largest prime divisor of $n$, unless $k=1$.


                                  Hence the conjecture fails only for $n=2$ (and is meaningless for $n=1$).






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    If the first of the $n$ summands is $m+1$, then the sum is
                                    $$(m+1)+(m+2)+ldots+(m+n)=nm+1+2+ldots+n=nm+frac{n(n+1)}{2}. $$




                                    • If $n$ is odd, say $n=2k-1$, this is even a multiple of $n$, namely $ncdot(m+k)$. Then even more so, it is a multiple of e.g. the largest prime divisor of $n$.

                                    • If $n$ is even, say $n=2k$, then it is at least a multiple of $k$, namely $kcdot(m+n+1)$. This is still a multiple of the largest prime divisor of $n$, unless $k=1$.


                                    Hence the conjecture fails only for $n=2$ (and is meaningless for $n=1$).






                                    share|cite|improve this answer









                                    $endgroup$



                                    If the first of the $n$ summands is $m+1$, then the sum is
                                    $$(m+1)+(m+2)+ldots+(m+n)=nm+1+2+ldots+n=nm+frac{n(n+1)}{2}. $$




                                    • If $n$ is odd, say $n=2k-1$, this is even a multiple of $n$, namely $ncdot(m+k)$. Then even more so, it is a multiple of e.g. the largest prime divisor of $n$.

                                    • If $n$ is even, say $n=2k$, then it is at least a multiple of $k$, namely $kcdot(m+n+1)$. This is still a multiple of the largest prime divisor of $n$, unless $k=1$.


                                    Hence the conjecture fails only for $n=2$ (and is meaningless for $n=1$).







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 47 mins ago









                                    Hagen von EitzenHagen von Eitzen

                                    279k23271503




                                    279k23271503






























                                        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%2f3112297%2fthe-sum-of-n-consecutive-numbers-is-divisible-by-the-greatest-prime-factor-of-n%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

                                        Plaza Victoria

                                        Image
                                        Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                                        Read more

                                        Puebla de Zaragoza

                                        Image
                                        Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                                        Read more

                                        Musa

                                        Image
                                        Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                                        Read more