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












1












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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.






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    1












    $begingroup$


    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









    $endgroup$















      1












      1








      1





      $begingroup$


      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






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      asked 41 mins ago









      MathGuyMathGuy

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          3 Answers
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          $begingroup$

          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









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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

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                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 30 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 22 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 21 mins ago









                                    Hagen von EitzenHagen von Eitzen

                                    279k23271503




                                    279k23271503






























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