What is the last digit of $n^5-5n^3+4n+7, forall n in mathbb{N}$? [closed]












2












$begingroup$



If $n$ is a natural number larger than two, what is the last digit of the expression $n^5-5n^3+4n+7$? How do you prove it ?







number-theory







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closed as off-topic by YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin Feb 12 at 11:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    What are your thoughts? Have you tried several $n$ at least?
    $endgroup$
    – lulu
    Feb 3 at 12:06






  • 1




    $begingroup$
    The last digit is $7$,maybe this helps you to come up with a proof.
    $endgroup$
    – Alexdanut
    Feb 3 at 12:11










  • $begingroup$
    yes i have already figure it out that the last digit is seven, however i cannot prove it.
    $endgroup$
    – dburanszki
    Feb 3 at 12:14






  • 2




    $begingroup$
    The assumption that $n>2$ is unnecessary. You get $7$ even for $n=0,1$.
    $endgroup$
    – Yanko
    Feb 3 at 12:20
















2












$begingroup$



If $n$ is a natural number larger than two, what is the last digit of the expression $n^5-5n^3+4n+7$? How do you prove it ?







number-theory







share|cite|improve this question











$endgroup$



closed as off-topic by YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin Feb 12 at 11:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    What are your thoughts? Have you tried several $n$ at least?
    $endgroup$
    – lulu
    Feb 3 at 12:06






  • 1




    $begingroup$
    The last digit is $7$,maybe this helps you to come up with a proof.
    $endgroup$
    – Alexdanut
    Feb 3 at 12:11










  • $begingroup$
    yes i have already figure it out that the last digit is seven, however i cannot prove it.
    $endgroup$
    – dburanszki
    Feb 3 at 12:14






  • 2




    $begingroup$
    The assumption that $n>2$ is unnecessary. You get $7$ even for $n=0,1$.
    $endgroup$
    – Yanko
    Feb 3 at 12:20














2












2








2


2



$begingroup$



If $n$ is a natural number larger than two, what is the last digit of the expression $n^5-5n^3+4n+7$? How do you prove it ?







number-theory







share|cite|improve this question











$endgroup$





If $n$ is a natural number larger than two, what is the last digit of the expression $n^5-5n^3+4n+7$? How do you prove it ?







number-theory




number-theory






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Feb 3 at 12:18









jvdhooft

5,65961641




5,65961641










asked Feb 3 at 12:05









dburanszkidburanszki

183




183




closed as off-topic by YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin Feb 12 at 11:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin Feb 12 at 11:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    What are your thoughts? Have you tried several $n$ at least?
    $endgroup$
    – lulu
    Feb 3 at 12:06






  • 1




    $begingroup$
    The last digit is $7$,maybe this helps you to come up with a proof.
    $endgroup$
    – Alexdanut
    Feb 3 at 12:11










  • $begingroup$
    yes i have already figure it out that the last digit is seven, however i cannot prove it.
    $endgroup$
    – dburanszki
    Feb 3 at 12:14






  • 2




    $begingroup$
    The assumption that $n>2$ is unnecessary. You get $7$ even for $n=0,1$.
    $endgroup$
    – Yanko
    Feb 3 at 12:20














  • 1




    $begingroup$
    What are your thoughts? Have you tried several $n$ at least?
    $endgroup$
    – lulu
    Feb 3 at 12:06






  • 1




    $begingroup$
    The last digit is $7$,maybe this helps you to come up with a proof.
    $endgroup$
    – Alexdanut
    Feb 3 at 12:11










  • $begingroup$
    yes i have already figure it out that the last digit is seven, however i cannot prove it.
    $endgroup$
    – dburanszki
    Feb 3 at 12:14






  • 2




    $begingroup$
    The assumption that $n>2$ is unnecessary. You get $7$ even for $n=0,1$.
    $endgroup$
    – Yanko
    Feb 3 at 12:20








1




1




$begingroup$
What are your thoughts? Have you tried several $n$ at least?
$endgroup$
– lulu
Feb 3 at 12:06




$begingroup$
What are your thoughts? Have you tried several $n$ at least?
$endgroup$
– lulu
Feb 3 at 12:06




1




1




$begingroup$
The last digit is $7$,maybe this helps you to come up with a proof.
$endgroup$
– Alexdanut
Feb 3 at 12:11




$begingroup$
The last digit is $7$,maybe this helps you to come up with a proof.
$endgroup$
– Alexdanut
Feb 3 at 12:11












$begingroup$
yes i have already figure it out that the last digit is seven, however i cannot prove it.
$endgroup$
– dburanszki
Feb 3 at 12:14




$begingroup$
yes i have already figure it out that the last digit is seven, however i cannot prove it.
$endgroup$
– dburanszki
Feb 3 at 12:14




2




2




$begingroup$
The assumption that $n>2$ is unnecessary. You get $7$ even for $n=0,1$.
$endgroup$
– Yanko
Feb 3 at 12:20




$begingroup$
The assumption that $n>2$ is unnecessary. You get $7$ even for $n=0,1$.
$endgroup$
– Yanko
Feb 3 at 12:20










5 Answers
5






active

oldest

votes


















2












$begingroup$

If $n in mathbb{N}$ we can try several values to see that the digit of the expression seems to always be $7$. So we need to prove that:
$$n^5−5n^3+4n+7 equiv 7 mod{10}$$
$$n^5−5n^3+4n equiv 0 mod{10}$$
$$n(n^4−5n^2+4) equiv 0 mod{10}$$
$$n(n^2-1)(n^2-4) equiv 0 mod{10}$$
$$n(n+1)(n-1)(n+2)(n-2) equiv 0 mod{10}$$
$$(n-2)(n-1)n(n+1)(n+2) equiv 0 mod{10}$$



And the above is true for all $n in mathbb{N}$ as $5$ consecutive numbers must contain a number divisible by $2$ and one divisible by $5$. Hence $10|(n-2)(n-1)n(n+1)(n+2)$ for all $n in mathbb{N}$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Hint: Your term can be written in the form $$7+(n-2)(n-1)n(n+1)(n+2)$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      No! 25 seconds before me...
      $endgroup$
      – Peter Foreman
      Feb 3 at 12:14



















    2












    $begingroup$

    Need to compute $n^5-5n^3+4n+7 $ (mod $10$). By Fermat's little theorem, $n^5-5n^3+4n+7equiv n-5n^3+4n+7equiv 2 $ (mod $5$), and $n^5-5n^3+4n+7equiv n-5n+4n+7equiv 1 $ (mod $2$), hence $n^5-5n^3+4n+7equiv 7 $ (mod $10$).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Really nice one.
      $endgroup$
      – Yanko
      Feb 3 at 12:19



















    1












    $begingroup$

    Let $n$ be a number whose last digit is $din{0,1,2,...,9}$.



    Then the last digit of $n^5 - 5n^3+4n^2+7$ is the last digit of $d^5-5d^3+4d^2+7$.



    So you only need to check these $10$ cases.



    Another option: Note that $n^5-5n^3+4n^2 = (n-2)(n-1)n(n+1)(n+2)$ and so it has to be divisible by $10$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Another variant:



      $$begin{array}{l}n^5-5n^3+4nequiv n^5+4nequiv n-n=0mod5\
      n^5-5n^3+4nequiv n^5-n^3equiv n-n=0mod 2
      end{array}biggr} quadtext{by } textit{ lil' Fermat}, $$



      so $;n^5-5n^3+4nequiv 0mod 10$, and ultimately
      $$n^5-5n^3+4n+7equiv 7mod 10. $$






      share|cite|improve this answer











      $endgroup$




















        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        If $n in mathbb{N}$ we can try several values to see that the digit of the expression seems to always be $7$. So we need to prove that:
        $$n^5−5n^3+4n+7 equiv 7 mod{10}$$
        $$n^5−5n^3+4n equiv 0 mod{10}$$
        $$n(n^4−5n^2+4) equiv 0 mod{10}$$
        $$n(n^2-1)(n^2-4) equiv 0 mod{10}$$
        $$n(n+1)(n-1)(n+2)(n-2) equiv 0 mod{10}$$
        $$(n-2)(n-1)n(n+1)(n+2) equiv 0 mod{10}$$



        And the above is true for all $n in mathbb{N}$ as $5$ consecutive numbers must contain a number divisible by $2$ and one divisible by $5$. Hence $10|(n-2)(n-1)n(n+1)(n+2)$ for all $n in mathbb{N}$.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          If $n in mathbb{N}$ we can try several values to see that the digit of the expression seems to always be $7$. So we need to prove that:
          $$n^5−5n^3+4n+7 equiv 7 mod{10}$$
          $$n^5−5n^3+4n equiv 0 mod{10}$$
          $$n(n^4−5n^2+4) equiv 0 mod{10}$$
          $$n(n^2-1)(n^2-4) equiv 0 mod{10}$$
          $$n(n+1)(n-1)(n+2)(n-2) equiv 0 mod{10}$$
          $$(n-2)(n-1)n(n+1)(n+2) equiv 0 mod{10}$$



          And the above is true for all $n in mathbb{N}$ as $5$ consecutive numbers must contain a number divisible by $2$ and one divisible by $5$. Hence $10|(n-2)(n-1)n(n+1)(n+2)$ for all $n in mathbb{N}$.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            If $n in mathbb{N}$ we can try several values to see that the digit of the expression seems to always be $7$. So we need to prove that:
            $$n^5−5n^3+4n+7 equiv 7 mod{10}$$
            $$n^5−5n^3+4n equiv 0 mod{10}$$
            $$n(n^4−5n^2+4) equiv 0 mod{10}$$
            $$n(n^2-1)(n^2-4) equiv 0 mod{10}$$
            $$n(n+1)(n-1)(n+2)(n-2) equiv 0 mod{10}$$
            $$(n-2)(n-1)n(n+1)(n+2) equiv 0 mod{10}$$



            And the above is true for all $n in mathbb{N}$ as $5$ consecutive numbers must contain a number divisible by $2$ and one divisible by $5$. Hence $10|(n-2)(n-1)n(n+1)(n+2)$ for all $n in mathbb{N}$.






            share|cite|improve this answer









            $endgroup$



            If $n in mathbb{N}$ we can try several values to see that the digit of the expression seems to always be $7$. So we need to prove that:
            $$n^5−5n^3+4n+7 equiv 7 mod{10}$$
            $$n^5−5n^3+4n equiv 0 mod{10}$$
            $$n(n^4−5n^2+4) equiv 0 mod{10}$$
            $$n(n^2-1)(n^2-4) equiv 0 mod{10}$$
            $$n(n+1)(n-1)(n+2)(n-2) equiv 0 mod{10}$$
            $$(n-2)(n-1)n(n+1)(n+2) equiv 0 mod{10}$$



            And the above is true for all $n in mathbb{N}$ as $5$ consecutive numbers must contain a number divisible by $2$ and one divisible by $5$. Hence $10|(n-2)(n-1)n(n+1)(n+2)$ for all $n in mathbb{N}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 3 at 12:13









            Peter ForemanPeter Foreman

            7,5361320




            7,5361320























                3












                $begingroup$

                Hint: Your term can be written in the form $$7+(n-2)(n-1)n(n+1)(n+2)$$






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  No! 25 seconds before me...
                  $endgroup$
                  – Peter Foreman
                  Feb 3 at 12:14
















                3












                $begingroup$

                Hint: Your term can be written in the form $$7+(n-2)(n-1)n(n+1)(n+2)$$






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  No! 25 seconds before me...
                  $endgroup$
                  – Peter Foreman
                  Feb 3 at 12:14














                3












                3








                3





                $begingroup$

                Hint: Your term can be written in the form $$7+(n-2)(n-1)n(n+1)(n+2)$$






                share|cite|improve this answer











                $endgroup$



                Hint: Your term can be written in the form $$7+(n-2)(n-1)n(n+1)(n+2)$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 3 at 12:15









                Yanko

                8,4692830




                8,4692830










                answered Feb 3 at 12:13









                Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                79k42867




                79k42867












                • $begingroup$
                  No! 25 seconds before me...
                  $endgroup$
                  – Peter Foreman
                  Feb 3 at 12:14


















                • $begingroup$
                  No! 25 seconds before me...
                  $endgroup$
                  – Peter Foreman
                  Feb 3 at 12:14
















                $begingroup$
                No! 25 seconds before me...
                $endgroup$
                – Peter Foreman
                Feb 3 at 12:14




                $begingroup$
                No! 25 seconds before me...
                $endgroup$
                – Peter Foreman
                Feb 3 at 12:14











                2












                $begingroup$

                Need to compute $n^5-5n^3+4n+7 $ (mod $10$). By Fermat's little theorem, $n^5-5n^3+4n+7equiv n-5n^3+4n+7equiv 2 $ (mod $5$), and $n^5-5n^3+4n+7equiv n-5n+4n+7equiv 1 $ (mod $2$), hence $n^5-5n^3+4n+7equiv 7 $ (mod $10$).






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Really nice one.
                  $endgroup$
                  – Yanko
                  Feb 3 at 12:19
















                2












                $begingroup$

                Need to compute $n^5-5n^3+4n+7 $ (mod $10$). By Fermat's little theorem, $n^5-5n^3+4n+7equiv n-5n^3+4n+7equiv 2 $ (mod $5$), and $n^5-5n^3+4n+7equiv n-5n+4n+7equiv 1 $ (mod $2$), hence $n^5-5n^3+4n+7equiv 7 $ (mod $10$).






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Really nice one.
                  $endgroup$
                  – Yanko
                  Feb 3 at 12:19














                2












                2








                2





                $begingroup$

                Need to compute $n^5-5n^3+4n+7 $ (mod $10$). By Fermat's little theorem, $n^5-5n^3+4n+7equiv n-5n^3+4n+7equiv 2 $ (mod $5$), and $n^5-5n^3+4n+7equiv n-5n+4n+7equiv 1 $ (mod $2$), hence $n^5-5n^3+4n+7equiv 7 $ (mod $10$).






                share|cite|improve this answer









                $endgroup$



                Need to compute $n^5-5n^3+4n+7 $ (mod $10$). By Fermat's little theorem, $n^5-5n^3+4n+7equiv n-5n^3+4n+7equiv 2 $ (mod $5$), and $n^5-5n^3+4n+7equiv n-5n+4n+7equiv 1 $ (mod $2$), hence $n^5-5n^3+4n+7equiv 7 $ (mod $10$).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 3 at 12:18









                BonbonBonbon

                45118




                45118












                • $begingroup$
                  Really nice one.
                  $endgroup$
                  – Yanko
                  Feb 3 at 12:19


















                • $begingroup$
                  Really nice one.
                  $endgroup$
                  – Yanko
                  Feb 3 at 12:19
















                $begingroup$
                Really nice one.
                $endgroup$
                – Yanko
                Feb 3 at 12:19




                $begingroup$
                Really nice one.
                $endgroup$
                – Yanko
                Feb 3 at 12:19











                1












                $begingroup$

                Let $n$ be a number whose last digit is $din{0,1,2,...,9}$.



                Then the last digit of $n^5 - 5n^3+4n^2+7$ is the last digit of $d^5-5d^3+4d^2+7$.



                So you only need to check these $10$ cases.



                Another option: Note that $n^5-5n^3+4n^2 = (n-2)(n-1)n(n+1)(n+2)$ and so it has to be divisible by $10$.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Let $n$ be a number whose last digit is $din{0,1,2,...,9}$.



                  Then the last digit of $n^5 - 5n^3+4n^2+7$ is the last digit of $d^5-5d^3+4d^2+7$.



                  So you only need to check these $10$ cases.



                  Another option: Note that $n^5-5n^3+4n^2 = (n-2)(n-1)n(n+1)(n+2)$ and so it has to be divisible by $10$.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Let $n$ be a number whose last digit is $din{0,1,2,...,9}$.



                    Then the last digit of $n^5 - 5n^3+4n^2+7$ is the last digit of $d^5-5d^3+4d^2+7$.



                    So you only need to check these $10$ cases.



                    Another option: Note that $n^5-5n^3+4n^2 = (n-2)(n-1)n(n+1)(n+2)$ and so it has to be divisible by $10$.






                    share|cite|improve this answer









                    $endgroup$



                    Let $n$ be a number whose last digit is $din{0,1,2,...,9}$.



                    Then the last digit of $n^5 - 5n^3+4n^2+7$ is the last digit of $d^5-5d^3+4d^2+7$.



                    So you only need to check these $10$ cases.



                    Another option: Note that $n^5-5n^3+4n^2 = (n-2)(n-1)n(n+1)(n+2)$ and so it has to be divisible by $10$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Feb 3 at 12:11









                    YankoYanko

                    8,4692830




                    8,4692830























                        1












                        $begingroup$

                        Another variant:



                        $$begin{array}{l}n^5-5n^3+4nequiv n^5+4nequiv n-n=0mod5\
                        n^5-5n^3+4nequiv n^5-n^3equiv n-n=0mod 2
                        end{array}biggr} quadtext{by } textit{ lil' Fermat}, $$



                        so $;n^5-5n^3+4nequiv 0mod 10$, and ultimately
                        $$n^5-5n^3+4n+7equiv 7mod 10. $$






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          Another variant:



                          $$begin{array}{l}n^5-5n^3+4nequiv n^5+4nequiv n-n=0mod5\
                          n^5-5n^3+4nequiv n^5-n^3equiv n-n=0mod 2
                          end{array}biggr} quadtext{by } textit{ lil' Fermat}, $$



                          so $;n^5-5n^3+4nequiv 0mod 10$, and ultimately
                          $$n^5-5n^3+4n+7equiv 7mod 10. $$






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Another variant:



                            $$begin{array}{l}n^5-5n^3+4nequiv n^5+4nequiv n-n=0mod5\
                            n^5-5n^3+4nequiv n^5-n^3equiv n-n=0mod 2
                            end{array}biggr} quadtext{by } textit{ lil' Fermat}, $$



                            so $;n^5-5n^3+4nequiv 0mod 10$, and ultimately
                            $$n^5-5n^3+4n+7equiv 7mod 10. $$






                            share|cite|improve this answer











                            $endgroup$



                            Another variant:



                            $$begin{array}{l}n^5-5n^3+4nequiv n^5+4nequiv n-n=0mod5\
                            n^5-5n^3+4nequiv n^5-n^3equiv n-n=0mod 2
                            end{array}biggr} quadtext{by } textit{ lil' Fermat}, $$



                            so $;n^5-5n^3+4nequiv 0mod 10$, and ultimately
                            $$n^5-5n^3+4n+7equiv 7mod 10. $$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Feb 3 at 12:47

























                            answered Feb 3 at 12:41









                            BernardBernard

                            124k742117




                            124k742117















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                                Setting WPF background Opacity programmatically

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                                .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 1 I'm working on application which creates a new wpf border component for each row in a database. This means I've got to style the border component in C# rather than XAML (as far as I'm aware). The styling is all good so far apart from trying to set the background opacity. motherboards.Add(new Border()); Border moboBorder = motherboards[i]; moboBorder.Width = 150; moboBorder.Height = 150; moboBorder.BorderBrush = Brushes.Black; moboBorder.Background = Brushes.White; moboBorder.CornerRadius = new CornerRadius(10); moboBorder.Margin = new Thickness(5); moboBorder.BorderThickness = new Thickness(1); You can a...
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                                in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

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                                1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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