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












2












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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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$endgroup$

















    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.










    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.










      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






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      asked 1 hour ago









      MathGuyMathGuy

      913315




      913315






















          4 Answers
          4






          active

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          5












          $begingroup$

          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









              $endgroup$





















                0












                $begingroup$

                One simple way of seeing it is the sum of $n $ consecutive integers is $n $ times the average of the consecutive integers.



                If $n $ is odd then the average of the consecutive integers is the middle integer and ... is an integer... so the sum is an integer times $n $



                If $n $ is even then the average of the consecutive integers is this mid point between two integers. This is an integer if multiply by $2$ so the sum is So the sum is an integer times $frac n2$ which is an integer..



                So long as $n>2$ ($n =2$ is a simple oversight; $1$ divides $2$ but.....) this stronger result necessitates the student's result as $n $ and $frac n2$ are divisible by all the primes except maybe $2$ (and $2$ is only a factor if $n $ is even and then $2$ isn't the largest prime unless $n $ is a power of $2$ and then $2$ does divide unless $n =2^1$ and then... oops.).






                share|cite|improve this answer









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                  4 Answers
                  4






                  active

                  oldest

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                  4 Answers
                  4






                  active

                  oldest

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                  active

                  oldest

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                  active

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                  5












                  $begingroup$

                  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









                  $endgroup$


















                    5












                    $begingroup$

                    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









                    $endgroup$
















                      5












                      5








                      5





                      $begingroup$

                      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 1 hour 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 1 hour 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 1 hour ago









                                      Hagen von EitzenHagen von Eitzen

                                      279k23271503




                                      279k23271503























                                          0












                                          $begingroup$

                                          One simple way of seeing it is the sum of $n $ consecutive integers is $n $ times the average of the consecutive integers.



                                          If $n $ is odd then the average of the consecutive integers is the middle integer and ... is an integer... so the sum is an integer times $n $



                                          If $n $ is even then the average of the consecutive integers is this mid point between two integers. This is an integer if multiply by $2$ so the sum is So the sum is an integer times $frac n2$ which is an integer..



                                          So long as $n>2$ ($n =2$ is a simple oversight; $1$ divides $2$ but.....) this stronger result necessitates the student's result as $n $ and $frac n2$ are divisible by all the primes except maybe $2$ (and $2$ is only a factor if $n $ is even and then $2$ isn't the largest prime unless $n $ is a power of $2$ and then $2$ does divide unless $n =2^1$ and then... oops.).






                                          share|cite|improve this answer









                                          $endgroup$


















                                            0












                                            $begingroup$

                                            One simple way of seeing it is the sum of $n $ consecutive integers is $n $ times the average of the consecutive integers.



                                            If $n $ is odd then the average of the consecutive integers is the middle integer and ... is an integer... so the sum is an integer times $n $



                                            If $n $ is even then the average of the consecutive integers is this mid point between two integers. This is an integer if multiply by $2$ so the sum is So the sum is an integer times $frac n2$ which is an integer..



                                            So long as $n>2$ ($n =2$ is a simple oversight; $1$ divides $2$ but.....) this stronger result necessitates the student's result as $n $ and $frac n2$ are divisible by all the primes except maybe $2$ (and $2$ is only a factor if $n $ is even and then $2$ isn't the largest prime unless $n $ is a power of $2$ and then $2$ does divide unless $n =2^1$ and then... oops.).






                                            share|cite|improve this answer









                                            $endgroup$
















                                              0












                                              0








                                              0





                                              $begingroup$

                                              One simple way of seeing it is the sum of $n $ consecutive integers is $n $ times the average of the consecutive integers.



                                              If $n $ is odd then the average of the consecutive integers is the middle integer and ... is an integer... so the sum is an integer times $n $



                                              If $n $ is even then the average of the consecutive integers is this mid point between two integers. This is an integer if multiply by $2$ so the sum is So the sum is an integer times $frac n2$ which is an integer..



                                              So long as $n>2$ ($n =2$ is a simple oversight; $1$ divides $2$ but.....) this stronger result necessitates the student's result as $n $ and $frac n2$ are divisible by all the primes except maybe $2$ (and $2$ is only a factor if $n $ is even and then $2$ isn't the largest prime unless $n $ is a power of $2$ and then $2$ does divide unless $n =2^1$ and then... oops.).






                                              share|cite|improve this answer









                                              $endgroup$



                                              One simple way of seeing it is the sum of $n $ consecutive integers is $n $ times the average of the consecutive integers.



                                              If $n $ is odd then the average of the consecutive integers is the middle integer and ... is an integer... so the sum is an integer times $n $



                                              If $n $ is even then the average of the consecutive integers is this mid point between two integers. This is an integer if multiply by $2$ so the sum is So the sum is an integer times $frac n2$ which is an integer..



                                              So long as $n>2$ ($n =2$ is a simple oversight; $1$ divides $2$ but.....) this stronger result necessitates the student's result as $n $ and $frac n2$ are divisible by all the primes except maybe $2$ (and $2$ is only a factor if $n $ is even and then $2$ isn't the largest prime unless $n $ is a power of $2$ and then $2$ does divide unless $n =2^1$ and then... oops.).







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered 19 mins ago









                                              fleabloodfleablood

                                              70.8k22686




                                              70.8k22686






























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