Factorising $99999,00000,99999,00001$












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I am reading an article in The Mathematical Gazette about factorising the 20 digit number
$$N=99999,00000,99999,00001.$$



It is stated that $N=dfrac{10^{25}+1}{10^5+1}$ (I understand why this is true), and consequently, if $p$ is a prime factor of $N$, then $p$ must be of the form $50k+1$. Why does this follow?










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    2












    $begingroup$


    I am reading an article in The Mathematical Gazette about factorising the 20 digit number
    $$N=99999,00000,99999,00001.$$



    It is stated that $N=dfrac{10^{25}+1}{10^5+1}$ (I understand why this is true), and consequently, if $p$ is a prime factor of $N$, then $p$ must be of the form $50k+1$. Why does this follow?










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I am reading an article in The Mathematical Gazette about factorising the 20 digit number
      $$N=99999,00000,99999,00001.$$



      It is stated that $N=dfrac{10^{25}+1}{10^5+1}$ (I understand why this is true), and consequently, if $p$ is a prime factor of $N$, then $p$ must be of the form $50k+1$. Why does this follow?










      share|cite|improve this question









      $endgroup$




      I am reading an article in The Mathematical Gazette about factorising the 20 digit number
      $$N=99999,00000,99999,00001.$$



      It is stated that $N=dfrac{10^{25}+1}{10^5+1}$ (I understand why this is true), and consequently, if $p$ is a prime factor of $N$, then $p$ must be of the form $50k+1$. Why does this follow?







      elementary-number-theory






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      asked Dec 20 '18 at 10:31









      rbirdrbird

      1,20414




      1,20414






















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          $begingroup$

          If $p$ is a prime factor of $N$, then it is a factor of $10^{25}+1$ and thus $$10^{25}equiv -1 implies 10^{50}equiv 1pmod p,$$
          i.e. if $d$ is the order of $10$ modulo $p$, then $d$ divides $50$. But $d$ cannot divide $25$ (otherwise we would have $10^{25}equiv 1pmod p$), and $d$ cannot divide $10$ either (otherwise $10^{10}equiv 1pmod p$, but one can easily verify that $gcd(N,10^{10}-1)=1$).



          Therefore, $d=50$. And Little Fermat says that $dmid(p-1)$.






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            $begingroup$

            If $p$ is a prime factor of $N$, then it is a factor of $10^{25}+1$ and thus $$10^{25}equiv -1 implies 10^{50}equiv 1pmod p,$$
            i.e. if $d$ is the order of $10$ modulo $p$, then $d$ divides $50$. But $d$ cannot divide $25$ (otherwise we would have $10^{25}equiv 1pmod p$), and $d$ cannot divide $10$ either (otherwise $10^{10}equiv 1pmod p$, but one can easily verify that $gcd(N,10^{10}-1)=1$).



            Therefore, $d=50$. And Little Fermat says that $dmid(p-1)$.






            share|cite|improve this answer









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              $begingroup$

              If $p$ is a prime factor of $N$, then it is a factor of $10^{25}+1$ and thus $$10^{25}equiv -1 implies 10^{50}equiv 1pmod p,$$
              i.e. if $d$ is the order of $10$ modulo $p$, then $d$ divides $50$. But $d$ cannot divide $25$ (otherwise we would have $10^{25}equiv 1pmod p$), and $d$ cannot divide $10$ either (otherwise $10^{10}equiv 1pmod p$, but one can easily verify that $gcd(N,10^{10}-1)=1$).



              Therefore, $d=50$. And Little Fermat says that $dmid(p-1)$.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                If $p$ is a prime factor of $N$, then it is a factor of $10^{25}+1$ and thus $$10^{25}equiv -1 implies 10^{50}equiv 1pmod p,$$
                i.e. if $d$ is the order of $10$ modulo $p$, then $d$ divides $50$. But $d$ cannot divide $25$ (otherwise we would have $10^{25}equiv 1pmod p$), and $d$ cannot divide $10$ either (otherwise $10^{10}equiv 1pmod p$, but one can easily verify that $gcd(N,10^{10}-1)=1$).



                Therefore, $d=50$. And Little Fermat says that $dmid(p-1)$.






                share|cite|improve this answer









                $endgroup$



                If $p$ is a prime factor of $N$, then it is a factor of $10^{25}+1$ and thus $$10^{25}equiv -1 implies 10^{50}equiv 1pmod p,$$
                i.e. if $d$ is the order of $10$ modulo $p$, then $d$ divides $50$. But $d$ cannot divide $25$ (otherwise we would have $10^{25}equiv 1pmod p$), and $d$ cannot divide $10$ either (otherwise $10^{10}equiv 1pmod p$, but one can easily verify that $gcd(N,10^{10}-1)=1$).



                Therefore, $d=50$. And Little Fermat says that $dmid(p-1)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 20 '18 at 11:13









                metamorphymetamorphy

                3,6821621




                3,6821621






























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