Does $p^n$ divide $(p^n-1)!$?












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Wilson's Theorem implies that if $p$ is a prime, then $p$ does not divide $(p-1)!$. Can we say something whether $p^n$ divides or not $(p^n-1)!$, where $n$ is any integer at least 3? What about $n=2$?






elementary-number-theory







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$endgroup$








  • 3




    $begingroup$
    This is almost trivial if you just write out the factorial and count the multiples of $p$, $p^2$, and so on.
    $endgroup$
    – Arthur
    Dec 11 '18 at 16:43










  • $begingroup$
    What do you think. All number less than $p^n-1$ divide ($p^n-1)!$ so $p|(p^n-1)!$ and $p^2|(p^n-1)$ and $p^3|(p^n-1)$.....
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:43






  • 1




    $begingroup$
    Not true for $n = 1$
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:44






  • 1




    $begingroup$
    @fleablood Or $n=p=2$.
    $endgroup$
    – Arthur
    Dec 11 '18 at 17:09
















0












$begingroup$


Wilson's Theorem implies that if $p$ is a prime, then $p$ does not divide $(p-1)!$. Can we say something whether $p^n$ divides or not $(p^n-1)!$, where $n$ is any integer at least 3? What about $n=2$?






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    This is almost trivial if you just write out the factorial and count the multiples of $p$, $p^2$, and so on.
    $endgroup$
    – Arthur
    Dec 11 '18 at 16:43










  • $begingroup$
    What do you think. All number less than $p^n-1$ divide ($p^n-1)!$ so $p|(p^n-1)!$ and $p^2|(p^n-1)$ and $p^3|(p^n-1)$.....
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:43






  • 1




    $begingroup$
    Not true for $n = 1$
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:44






  • 1




    $begingroup$
    @fleablood Or $n=p=2$.
    $endgroup$
    – Arthur
    Dec 11 '18 at 17:09














0












0








0





$begingroup$


Wilson's Theorem implies that if $p$ is a prime, then $p$ does not divide $(p-1)!$. Can we say something whether $p^n$ divides or not $(p^n-1)!$, where $n$ is any integer at least 3? What about $n=2$?






elementary-number-theory







share|cite|improve this question











$endgroup$




Wilson's Theorem implies that if $p$ is a prime, then $p$ does not divide $(p-1)!$. Can we say something whether $p^n$ divides or not $(p^n-1)!$, where $n$ is any integer at least 3? What about $n=2$?






elementary-number-theory




elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 16:46









Mike

3,785411




3,785411










asked Dec 11 '18 at 16:39









LucaLuca

18119




18119








  • 3




    $begingroup$
    This is almost trivial if you just write out the factorial and count the multiples of $p$, $p^2$, and so on.
    $endgroup$
    – Arthur
    Dec 11 '18 at 16:43










  • $begingroup$
    What do you think. All number less than $p^n-1$ divide ($p^n-1)!$ so $p|(p^n-1)!$ and $p^2|(p^n-1)$ and $p^3|(p^n-1)$.....
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:43






  • 1




    $begingroup$
    Not true for $n = 1$
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:44






  • 1




    $begingroup$
    @fleablood Or $n=p=2$.
    $endgroup$
    – Arthur
    Dec 11 '18 at 17:09














  • 3




    $begingroup$
    This is almost trivial if you just write out the factorial and count the multiples of $p$, $p^2$, and so on.
    $endgroup$
    – Arthur
    Dec 11 '18 at 16:43










  • $begingroup$
    What do you think. All number less than $p^n-1$ divide ($p^n-1)!$ so $p|(p^n-1)!$ and $p^2|(p^n-1)$ and $p^3|(p^n-1)$.....
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:43






  • 1




    $begingroup$
    Not true for $n = 1$
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:44






  • 1




    $begingroup$
    @fleablood Or $n=p=2$.
    $endgroup$
    – Arthur
    Dec 11 '18 at 17:09








3




3




$begingroup$
This is almost trivial if you just write out the factorial and count the multiples of $p$, $p^2$, and so on.
$endgroup$
– Arthur
Dec 11 '18 at 16:43




$begingroup$
This is almost trivial if you just write out the factorial and count the multiples of $p$, $p^2$, and so on.
$endgroup$
– Arthur
Dec 11 '18 at 16:43












$begingroup$
What do you think. All number less than $p^n-1$ divide ($p^n-1)!$ so $p|(p^n-1)!$ and $p^2|(p^n-1)$ and $p^3|(p^n-1)$.....
$endgroup$
– fleablood
Dec 11 '18 at 16:43




$begingroup$
What do you think. All number less than $p^n-1$ divide ($p^n-1)!$ so $p|(p^n-1)!$ and $p^2|(p^n-1)$ and $p^3|(p^n-1)$.....
$endgroup$
– fleablood
Dec 11 '18 at 16:43




1




1




$begingroup$
Not true for $n = 1$
$endgroup$
– fleablood
Dec 11 '18 at 16:44




$begingroup$
Not true for $n = 1$
$endgroup$
– fleablood
Dec 11 '18 at 16:44




1




1




$begingroup$
@fleablood Or $n=p=2$.
$endgroup$
– Arthur
Dec 11 '18 at 17:09




$begingroup$
@fleablood Or $n=p=2$.
$endgroup$
– Arthur
Dec 11 '18 at 17:09










2 Answers
2






active

oldest

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1












$begingroup$

$(p^n - 1)! = 1*2*3......*(p^n-1)$ so



Note the $p, p^2, p^3,.....,p^{n-1}$ are all less than $p^n-1$.



Also $p, 2p, 3p, .....,p^n-p$ are also less than $p^n-1$.



So $p*2p*3p*...p^2...etc$ will divide $(p^n-1)!$.



In particular $p*p^2*p^3*....*p^{n-1} = p^{1+2+...(n-1)} = p^{frac {(n-1)n}2}$ will divide $(p^n-1)!$.



(Note: If $p> 2$ then this will not be the largest power of $p$ to divide $(p^n-1)!$ because we are not taking $2p*3p*...*(p-1)p*(p+1)p...$ etc into account. We'll deal with those later.)



So this will be true whenever $frac {n-1}2n ge n$ or if $nge 3$.



Okay, be what if $n = 2$ or $n=1$? Well obviously if $n =1$ we $(p-1)! = 1*2*....*(p-1)$ and $p$ doesn't divide any $k < p$ so the statement is false for $n=1$.



So what about $n =2$? Well now is when we'll take the $p, 2p, 3p...(p-1)p$ into account.



$(p^2-1)! = 1*2*3....*(p^2 - 1)$ and so $p, 2p,...., (p-1)p$ are all less than $p^2 -1$ so $p*2p*....*(p-1)p = (p-1)!p^{p-1}$ will divide $(p^2 -1)!$.



So if $p-1 ge 2$ or in other words if $p ge 3$ then the will be true for $n=2$.



And if $p =2$ and $n=2$? Well, then it is false. $(2^2-1)!= 1*color{blue}2*(2^2-1)$ and that's not enough copies of $color{blue}2$.



So conclusion:



False for $n =1$. False for $n = 2;p=2$. And true for all other cases.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    It is a well-known and easy-to-prove fact that the multiplicity with which a prime $p$ divides $n!$ is equal to
    $$sum_{k=1}^infty bigglfloor frac{n}{p^k}biggrfloor$$
    Thus, the multiplicity with which $p$ divides $(p^n-1)!$ is equal to
    $$sum_{k=1}^infty bigglfloor frac{p^n-1}{p^k}biggrfloor=sum_{k=1}^n (p^{n-k}-1)=frac{p^n-1}{p-1}-n$$
    Which is greater than $n$, for reasonably large $n$ (since it grows exponentially). Thus, for "sufficiently large" $n$, we have that $p^n|(p^n-1)!$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      how did you move from $sum_{k=1}^{infty} bigglfloor frac{p^n-1}{p^k} biggrfloor$ to $sum_{k=1}^n (p^{n-k}-1)$? should it be $sum_{k=1}^n (p^{n-k})$ as $bigglfloor frac{1}{p^k}biggrfloor = 0$
      $endgroup$
      – Maged Saeed
      Dec 11 '18 at 22:12












    • $begingroup$
      @MagedSaeed No. Notice that $$bigglfloorfrac{p^n-1}{p^k}biggrfloor=lfloor p^{n-k}-1/p^krfloor$$ and that if $xinmathbb Z$ and $ylt 1$, then $lfloor x-yrfloor=x-1$.
      $endgroup$
      – Frpzzd
      Dec 11 '18 at 23:21










    • $begingroup$
      Oh, I see. Thanks,
      $endgroup$
      – Maged Saeed
      Dec 12 '18 at 1:36











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    2 Answers
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    2 Answers
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    active

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    1












    $begingroup$

    $(p^n - 1)! = 1*2*3......*(p^n-1)$ so



    Note the $p, p^2, p^3,.....,p^{n-1}$ are all less than $p^n-1$.



    Also $p, 2p, 3p, .....,p^n-p$ are also less than $p^n-1$.



    So $p*2p*3p*...p^2...etc$ will divide $(p^n-1)!$.



    In particular $p*p^2*p^3*....*p^{n-1} = p^{1+2+...(n-1)} = p^{frac {(n-1)n}2}$ will divide $(p^n-1)!$.



    (Note: If $p> 2$ then this will not be the largest power of $p$ to divide $(p^n-1)!$ because we are not taking $2p*3p*...*(p-1)p*(p+1)p...$ etc into account. We'll deal with those later.)



    So this will be true whenever $frac {n-1}2n ge n$ or if $nge 3$.



    Okay, be what if $n = 2$ or $n=1$? Well obviously if $n =1$ we $(p-1)! = 1*2*....*(p-1)$ and $p$ doesn't divide any $k < p$ so the statement is false for $n=1$.



    So what about $n =2$? Well now is when we'll take the $p, 2p, 3p...(p-1)p$ into account.



    $(p^2-1)! = 1*2*3....*(p^2 - 1)$ and so $p, 2p,...., (p-1)p$ are all less than $p^2 -1$ so $p*2p*....*(p-1)p = (p-1)!p^{p-1}$ will divide $(p^2 -1)!$.



    So if $p-1 ge 2$ or in other words if $p ge 3$ then the will be true for $n=2$.



    And if $p =2$ and $n=2$? Well, then it is false. $(2^2-1)!= 1*color{blue}2*(2^2-1)$ and that's not enough copies of $color{blue}2$.



    So conclusion:



    False for $n =1$. False for $n = 2;p=2$. And true for all other cases.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      $(p^n - 1)! = 1*2*3......*(p^n-1)$ so



      Note the $p, p^2, p^3,.....,p^{n-1}$ are all less than $p^n-1$.



      Also $p, 2p, 3p, .....,p^n-p$ are also less than $p^n-1$.



      So $p*2p*3p*...p^2...etc$ will divide $(p^n-1)!$.



      In particular $p*p^2*p^3*....*p^{n-1} = p^{1+2+...(n-1)} = p^{frac {(n-1)n}2}$ will divide $(p^n-1)!$.



      (Note: If $p> 2$ then this will not be the largest power of $p$ to divide $(p^n-1)!$ because we are not taking $2p*3p*...*(p-1)p*(p+1)p...$ etc into account. We'll deal with those later.)



      So this will be true whenever $frac {n-1}2n ge n$ or if $nge 3$.



      Okay, be what if $n = 2$ or $n=1$? Well obviously if $n =1$ we $(p-1)! = 1*2*....*(p-1)$ and $p$ doesn't divide any $k < p$ so the statement is false for $n=1$.



      So what about $n =2$? Well now is when we'll take the $p, 2p, 3p...(p-1)p$ into account.



      $(p^2-1)! = 1*2*3....*(p^2 - 1)$ and so $p, 2p,...., (p-1)p$ are all less than $p^2 -1$ so $p*2p*....*(p-1)p = (p-1)!p^{p-1}$ will divide $(p^2 -1)!$.



      So if $p-1 ge 2$ or in other words if $p ge 3$ then the will be true for $n=2$.



      And if $p =2$ and $n=2$? Well, then it is false. $(2^2-1)!= 1*color{blue}2*(2^2-1)$ and that's not enough copies of $color{blue}2$.



      So conclusion:



      False for $n =1$. False for $n = 2;p=2$. And true for all other cases.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        $(p^n - 1)! = 1*2*3......*(p^n-1)$ so



        Note the $p, p^2, p^3,.....,p^{n-1}$ are all less than $p^n-1$.



        Also $p, 2p, 3p, .....,p^n-p$ are also less than $p^n-1$.



        So $p*2p*3p*...p^2...etc$ will divide $(p^n-1)!$.



        In particular $p*p^2*p^3*....*p^{n-1} = p^{1+2+...(n-1)} = p^{frac {(n-1)n}2}$ will divide $(p^n-1)!$.



        (Note: If $p> 2$ then this will not be the largest power of $p$ to divide $(p^n-1)!$ because we are not taking $2p*3p*...*(p-1)p*(p+1)p...$ etc into account. We'll deal with those later.)



        So this will be true whenever $frac {n-1}2n ge n$ or if $nge 3$.



        Okay, be what if $n = 2$ or $n=1$? Well obviously if $n =1$ we $(p-1)! = 1*2*....*(p-1)$ and $p$ doesn't divide any $k < p$ so the statement is false for $n=1$.



        So what about $n =2$? Well now is when we'll take the $p, 2p, 3p...(p-1)p$ into account.



        $(p^2-1)! = 1*2*3....*(p^2 - 1)$ and so $p, 2p,...., (p-1)p$ are all less than $p^2 -1$ so $p*2p*....*(p-1)p = (p-1)!p^{p-1}$ will divide $(p^2 -1)!$.



        So if $p-1 ge 2$ or in other words if $p ge 3$ then the will be true for $n=2$.



        And if $p =2$ and $n=2$? Well, then it is false. $(2^2-1)!= 1*color{blue}2*(2^2-1)$ and that's not enough copies of $color{blue}2$.



        So conclusion:



        False for $n =1$. False for $n = 2;p=2$. And true for all other cases.






        share|cite|improve this answer











        $endgroup$



        $(p^n - 1)! = 1*2*3......*(p^n-1)$ so



        Note the $p, p^2, p^3,.....,p^{n-1}$ are all less than $p^n-1$.



        Also $p, 2p, 3p, .....,p^n-p$ are also less than $p^n-1$.



        So $p*2p*3p*...p^2...etc$ will divide $(p^n-1)!$.



        In particular $p*p^2*p^3*....*p^{n-1} = p^{1+2+...(n-1)} = p^{frac {(n-1)n}2}$ will divide $(p^n-1)!$.



        (Note: If $p> 2$ then this will not be the largest power of $p$ to divide $(p^n-1)!$ because we are not taking $2p*3p*...*(p-1)p*(p+1)p...$ etc into account. We'll deal with those later.)



        So this will be true whenever $frac {n-1}2n ge n$ or if $nge 3$.



        Okay, be what if $n = 2$ or $n=1$? Well obviously if $n =1$ we $(p-1)! = 1*2*....*(p-1)$ and $p$ doesn't divide any $k < p$ so the statement is false for $n=1$.



        So what about $n =2$? Well now is when we'll take the $p, 2p, 3p...(p-1)p$ into account.



        $(p^2-1)! = 1*2*3....*(p^2 - 1)$ and so $p, 2p,...., (p-1)p$ are all less than $p^2 -1$ so $p*2p*....*(p-1)p = (p-1)!p^{p-1}$ will divide $(p^2 -1)!$.



        So if $p-1 ge 2$ or in other words if $p ge 3$ then the will be true for $n=2$.



        And if $p =2$ and $n=2$? Well, then it is false. $(2^2-1)!= 1*color{blue}2*(2^2-1)$ and that's not enough copies of $color{blue}2$.



        So conclusion:



        False for $n =1$. False for $n = 2;p=2$. And true for all other cases.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 '18 at 17:51

























        answered Dec 11 '18 at 17:08









        fleabloodfleablood

        69.7k22685




        69.7k22685























            3












            $begingroup$

            It is a well-known and easy-to-prove fact that the multiplicity with which a prime $p$ divides $n!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{n}{p^k}biggrfloor$$
            Thus, the multiplicity with which $p$ divides $(p^n-1)!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{p^n-1}{p^k}biggrfloor=sum_{k=1}^n (p^{n-k}-1)=frac{p^n-1}{p-1}-n$$
            Which is greater than $n$, for reasonably large $n$ (since it grows exponentially). Thus, for "sufficiently large" $n$, we have that $p^n|(p^n-1)!$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              how did you move from $sum_{k=1}^{infty} bigglfloor frac{p^n-1}{p^k} biggrfloor$ to $sum_{k=1}^n (p^{n-k}-1)$? should it be $sum_{k=1}^n (p^{n-k})$ as $bigglfloor frac{1}{p^k}biggrfloor = 0$
              $endgroup$
              – Maged Saeed
              Dec 11 '18 at 22:12












            • $begingroup$
              @MagedSaeed No. Notice that $$bigglfloorfrac{p^n-1}{p^k}biggrfloor=lfloor p^{n-k}-1/p^krfloor$$ and that if $xinmathbb Z$ and $ylt 1$, then $lfloor x-yrfloor=x-1$.
              $endgroup$
              – Frpzzd
              Dec 11 '18 at 23:21










            • $begingroup$
              Oh, I see. Thanks,
              $endgroup$
              – Maged Saeed
              Dec 12 '18 at 1:36
















            3












            $begingroup$

            It is a well-known and easy-to-prove fact that the multiplicity with which a prime $p$ divides $n!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{n}{p^k}biggrfloor$$
            Thus, the multiplicity with which $p$ divides $(p^n-1)!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{p^n-1}{p^k}biggrfloor=sum_{k=1}^n (p^{n-k}-1)=frac{p^n-1}{p-1}-n$$
            Which is greater than $n$, for reasonably large $n$ (since it grows exponentially). Thus, for "sufficiently large" $n$, we have that $p^n|(p^n-1)!$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              how did you move from $sum_{k=1}^{infty} bigglfloor frac{p^n-1}{p^k} biggrfloor$ to $sum_{k=1}^n (p^{n-k}-1)$? should it be $sum_{k=1}^n (p^{n-k})$ as $bigglfloor frac{1}{p^k}biggrfloor = 0$
              $endgroup$
              – Maged Saeed
              Dec 11 '18 at 22:12












            • $begingroup$
              @MagedSaeed No. Notice that $$bigglfloorfrac{p^n-1}{p^k}biggrfloor=lfloor p^{n-k}-1/p^krfloor$$ and that if $xinmathbb Z$ and $ylt 1$, then $lfloor x-yrfloor=x-1$.
              $endgroup$
              – Frpzzd
              Dec 11 '18 at 23:21










            • $begingroup$
              Oh, I see. Thanks,
              $endgroup$
              – Maged Saeed
              Dec 12 '18 at 1:36














            3












            3








            3





            $begingroup$

            It is a well-known and easy-to-prove fact that the multiplicity with which a prime $p$ divides $n!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{n}{p^k}biggrfloor$$
            Thus, the multiplicity with which $p$ divides $(p^n-1)!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{p^n-1}{p^k}biggrfloor=sum_{k=1}^n (p^{n-k}-1)=frac{p^n-1}{p-1}-n$$
            Which is greater than $n$, for reasonably large $n$ (since it grows exponentially). Thus, for "sufficiently large" $n$, we have that $p^n|(p^n-1)!$.






            share|cite|improve this answer









            $endgroup$



            It is a well-known and easy-to-prove fact that the multiplicity with which a prime $p$ divides $n!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{n}{p^k}biggrfloor$$
            Thus, the multiplicity with which $p$ divides $(p^n-1)!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{p^n-1}{p^k}biggrfloor=sum_{k=1}^n (p^{n-k}-1)=frac{p^n-1}{p-1}-n$$
            Which is greater than $n$, for reasonably large $n$ (since it grows exponentially). Thus, for "sufficiently large" $n$, we have that $p^n|(p^n-1)!$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 11 '18 at 16:45









            FrpzzdFrpzzd

            22.9k841109




            22.9k841109












            • $begingroup$
              how did you move from $sum_{k=1}^{infty} bigglfloor frac{p^n-1}{p^k} biggrfloor$ to $sum_{k=1}^n (p^{n-k}-1)$? should it be $sum_{k=1}^n (p^{n-k})$ as $bigglfloor frac{1}{p^k}biggrfloor = 0$
              $endgroup$
              – Maged Saeed
              Dec 11 '18 at 22:12












            • $begingroup$
              @MagedSaeed No. Notice that $$bigglfloorfrac{p^n-1}{p^k}biggrfloor=lfloor p^{n-k}-1/p^krfloor$$ and that if $xinmathbb Z$ and $ylt 1$, then $lfloor x-yrfloor=x-1$.
              $endgroup$
              – Frpzzd
              Dec 11 '18 at 23:21










            • $begingroup$
              Oh, I see. Thanks,
              $endgroup$
              – Maged Saeed
              Dec 12 '18 at 1:36


















            • $begingroup$
              how did you move from $sum_{k=1}^{infty} bigglfloor frac{p^n-1}{p^k} biggrfloor$ to $sum_{k=1}^n (p^{n-k}-1)$? should it be $sum_{k=1}^n (p^{n-k})$ as $bigglfloor frac{1}{p^k}biggrfloor = 0$
              $endgroup$
              – Maged Saeed
              Dec 11 '18 at 22:12












            • $begingroup$
              @MagedSaeed No. Notice that $$bigglfloorfrac{p^n-1}{p^k}biggrfloor=lfloor p^{n-k}-1/p^krfloor$$ and that if $xinmathbb Z$ and $ylt 1$, then $lfloor x-yrfloor=x-1$.
              $endgroup$
              – Frpzzd
              Dec 11 '18 at 23:21










            • $begingroup$
              Oh, I see. Thanks,
              $endgroup$
              – Maged Saeed
              Dec 12 '18 at 1:36
















            $begingroup$
            how did you move from $sum_{k=1}^{infty} bigglfloor frac{p^n-1}{p^k} biggrfloor$ to $sum_{k=1}^n (p^{n-k}-1)$? should it be $sum_{k=1}^n (p^{n-k})$ as $bigglfloor frac{1}{p^k}biggrfloor = 0$
            $endgroup$
            – Maged Saeed
            Dec 11 '18 at 22:12






            $begingroup$
            how did you move from $sum_{k=1}^{infty} bigglfloor frac{p^n-1}{p^k} biggrfloor$ to $sum_{k=1}^n (p^{n-k}-1)$? should it be $sum_{k=1}^n (p^{n-k})$ as $bigglfloor frac{1}{p^k}biggrfloor = 0$
            $endgroup$
            – Maged Saeed
            Dec 11 '18 at 22:12














            $begingroup$
            @MagedSaeed No. Notice that $$bigglfloorfrac{p^n-1}{p^k}biggrfloor=lfloor p^{n-k}-1/p^krfloor$$ and that if $xinmathbb Z$ and $ylt 1$, then $lfloor x-yrfloor=x-1$.
            $endgroup$
            – Frpzzd
            Dec 11 '18 at 23:21




            $begingroup$
            @MagedSaeed No. Notice that $$bigglfloorfrac{p^n-1}{p^k}biggrfloor=lfloor p^{n-k}-1/p^krfloor$$ and that if $xinmathbb Z$ and $ylt 1$, then $lfloor x-yrfloor=x-1$.
            $endgroup$
            – Frpzzd
            Dec 11 '18 at 23:21












            $begingroup$
            Oh, I see. Thanks,
            $endgroup$
            – Maged Saeed
            Dec 12 '18 at 1:36




            $begingroup$
            Oh, I see. Thanks,
            $endgroup$
            – Maged Saeed
            Dec 12 '18 at 1:36


















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