How to prove divisibility of a number using the binomial expansion?

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I have the following problem:



Prove that $6^n-1$ is always divisible by 5 using the binomial expansion of $(5+1)^n$.



How can I do this? I don't know how to begin, as I don't see how the binomial expansion relates to the question. Any help would be appreciated.






algebra-precalculus binomial-coefficients binomial-theorem







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  • See here for a better example and also a presentation using modular arithmetic.
    – Bill Dubuque
    1 hour ago















up vote
1
down vote

favorite












I have the following problem:



Prove that $6^n-1$ is always divisible by 5 using the binomial expansion of $(5+1)^n$.



How can I do this? I don't know how to begin, as I don't see how the binomial expansion relates to the question. Any help would be appreciated.






algebra-precalculus binomial-coefficients binomial-theorem







share|cite|improve this question





















  • See here for a better example and also a presentation using modular arithmetic.
    – Bill Dubuque
    1 hour ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have the following problem:



Prove that $6^n-1$ is always divisible by 5 using the binomial expansion of $(5+1)^n$.



How can I do this? I don't know how to begin, as I don't see how the binomial expansion relates to the question. Any help would be appreciated.






algebra-precalculus binomial-coefficients binomial-theorem







share|cite|improve this question













I have the following problem:



Prove that $6^n-1$ is always divisible by 5 using the binomial expansion of $(5+1)^n$.



How can I do this? I don't know how to begin, as I don't see how the binomial expansion relates to the question. Any help would be appreciated.






algebra-precalculus binomial-coefficients binomial-theorem




algebra-precalculus binomial-coefficients binomial-theorem






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









Adam Grey

294




294











  • See here for a better example and also a presentation using modular arithmetic.
    – Bill Dubuque
    1 hour ago

















  • See here for a better example and also a presentation using modular arithmetic.
    – Bill Dubuque
    1 hour ago
















See here for a better example and also a presentation using modular arithmetic.
– Bill Dubuque
1 hour ago





See here for a better example and also a presentation using modular arithmetic.
– Bill Dubuque
1 hour ago











4 Answers
4






active

oldest

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up vote
3
down vote



accepted










$$(5+1)^n-1 =underbracenchoose 05^n+ nchoose 15^n-1+...+ nchoose n-15_=5cdot (....)+1-1 = 5k$$






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    up vote
    2
    down vote













    HINT:
    $$(5+1)^n=sum_k=0^n binomnk5^k$$
    Which terms of this sum are divisible by $5$?






    share|cite|improve this answer




















    • (+1) I would have written an answer similar to this.
      – robjohn♦
      46 mins ago

















    up vote
    0
    down vote













    Guide:



    • Treat $a=5$ and $b=1$ in $(a+b)^n$, expand it using binomial expansion. Then subtract it by $1$.


    • Then try to factor $5$ out from the remaining term.






    share|cite|improve this answer



























      up vote
      0
      down vote













      $$6^n-1 = (5+1)^n-1$$
      $$(5+1)^n = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1$$
      $$(5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1-1$$
      $n choose n1 = 1$ so $1$ and $-1$ cancel out.
      $$implies (5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1$$
      All terms have a factor of $5$. Therefore, $6^n-1$ is divisible by $5$.






      share|cite|improve this answer






















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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

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        active

        oldest

        votes








        up vote
        3
        down vote



        accepted










        $$(5+1)^n-1 =underbracenchoose 05^n+ nchoose 15^n-1+...+ nchoose n-15_=5cdot (....)+1-1 = 5k$$






        share|cite|improve this answer
























          up vote
          3
          down vote



          accepted










          $$(5+1)^n-1 =underbracenchoose 05^n+ nchoose 15^n-1+...+ nchoose n-15_=5cdot (....)+1-1 = 5k$$






          share|cite|improve this answer






















            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            $$(5+1)^n-1 =underbracenchoose 05^n+ nchoose 15^n-1+...+ nchoose n-15_=5cdot (....)+1-1 = 5k$$






            share|cite|improve this answer












            $$(5+1)^n-1 =underbracenchoose 05^n+ nchoose 15^n-1+...+ nchoose n-15_=5cdot (....)+1-1 = 5k$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            greedoid

            30.7k94184




            30.7k94184




















                up vote
                2
                down vote













                HINT:
                $$(5+1)^n=sum_k=0^n binomnk5^k$$
                Which terms of this sum are divisible by $5$?






                share|cite|improve this answer




















                • (+1) I would have written an answer similar to this.
                  – robjohn♦
                  46 mins ago














                up vote
                2
                down vote













                HINT:
                $$(5+1)^n=sum_k=0^n binomnk5^k$$
                Which terms of this sum are divisible by $5$?






                share|cite|improve this answer




















                • (+1) I would have written an answer similar to this.
                  – robjohn♦
                  46 mins ago












                up vote
                2
                down vote










                up vote
                2
                down vote









                HINT:
                $$(5+1)^n=sum_k=0^n binomnk5^k$$
                Which terms of this sum are divisible by $5$?






                share|cite|improve this answer












                HINT:
                $$(5+1)^n=sum_k=0^n binomnk5^k$$
                Which terms of this sum are divisible by $5$?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Frpzzd

                18.1k63695




                18.1k63695











                • (+1) I would have written an answer similar to this.
                  – robjohn♦
                  46 mins ago
















                • (+1) I would have written an answer similar to this.
                  – robjohn♦
                  46 mins ago















                (+1) I would have written an answer similar to this.
                – robjohn♦
                46 mins ago




                (+1) I would have written an answer similar to this.
                – robjohn♦
                46 mins ago










                up vote
                0
                down vote













                Guide:



                • Treat $a=5$ and $b=1$ in $(a+b)^n$, expand it using binomial expansion. Then subtract it by $1$.


                • Then try to factor $5$ out from the remaining term.






                share|cite|improve this answer
























                  up vote
                  0
                  down vote













                  Guide:



                  • Treat $a=5$ and $b=1$ in $(a+b)^n$, expand it using binomial expansion. Then subtract it by $1$.


                  • Then try to factor $5$ out from the remaining term.






                  share|cite|improve this answer






















                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    Guide:



                    • Treat $a=5$ and $b=1$ in $(a+b)^n$, expand it using binomial expansion. Then subtract it by $1$.


                    • Then try to factor $5$ out from the remaining term.






                    share|cite|improve this answer












                    Guide:



                    • Treat $a=5$ and $b=1$ in $(a+b)^n$, expand it using binomial expansion. Then subtract it by $1$.


                    • Then try to factor $5$ out from the remaining term.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    Siong Thye Goh

                    85.1k1457107




                    85.1k1457107




















                        up vote
                        0
                        down vote













                        $$6^n-1 = (5+1)^n-1$$
                        $$(5+1)^n = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1$$
                        $$(5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1-1$$
                        $n choose n1 = 1$ so $1$ and $-1$ cancel out.
                        $$implies (5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1$$
                        All terms have a factor of $5$. Therefore, $6^n-1$ is divisible by $5$.






                        share|cite|improve this answer


























                          up vote
                          0
                          down vote













                          $$6^n-1 = (5+1)^n-1$$
                          $$(5+1)^n = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1$$
                          $$(5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1-1$$
                          $n choose n1 = 1$ so $1$ and $-1$ cancel out.
                          $$implies (5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1$$
                          All terms have a factor of $5$. Therefore, $6^n-1$ is divisible by $5$.






                          share|cite|improve this answer
























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            $$6^n-1 = (5+1)^n-1$$
                            $$(5+1)^n = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1$$
                            $$(5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1-1$$
                            $n choose n1 = 1$ so $1$ and $-1$ cancel out.
                            $$implies (5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1$$
                            All terms have a factor of $5$. Therefore, $6^n-1$ is divisible by $5$.






                            share|cite|improve this answer














                            $$6^n-1 = (5+1)^n-1$$
                            $$(5+1)^n = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1$$
                            $$(5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1-1$$
                            $n choose n1 = 1$ so $1$ and $-1$ cancel out.
                            $$implies (5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1$$
                            All terms have a factor of $5$. Therefore, $6^n-1$ is divisible by $5$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 1 hour ago

























                            answered 1 hour ago









                            KM101

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