About The Sum of Positive Divisors of $n$












3












$begingroup$


The question says:




Find the smallest positive integer $n$ so that $sigma(x)=n$ has no solution, exactly two solutions, exactly three solutions.




I could not come up with a good way to solve this question other that trial and error. But I am questioning this method. Is there any better ideas?






number-theory elementary-number-theory divisor-sum arithmetic-functions multiplicative-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think brute search is optimal, if a bit tedious. You'll find the answer fairly quickly.
    $endgroup$
    – lulu
    Dec 9 '18 at 13:35










  • $begingroup$
    Oh, thanks, my method sounds good though.
    $endgroup$
    – Maged Saeed
    Dec 9 '18 at 13:35






  • 1




    $begingroup$
    Absolutely, yes. And you are correct about $2,12$. The third part won't take you as long as you fear. Keep in mind $sigma_1(n)≥n+1$ with equality only for primes. That makes it easy to truncate your search.
    $endgroup$
    – lulu
    Dec 9 '18 at 13:36








  • 1




    $begingroup$
    Correct again. $,$
    $endgroup$
    – lulu
    Dec 9 '18 at 13:39






  • 1




    $begingroup$
    (+1) for the posted solution, good work.
    $endgroup$
    – lulu
    Dec 9 '18 at 13:49
















3












$begingroup$


The question says:




Find the smallest positive integer $n$ so that $sigma(x)=n$ has no solution, exactly two solutions, exactly three solutions.




I could not come up with a good way to solve this question other that trial and error. But I am questioning this method. Is there any better ideas?






number-theory elementary-number-theory divisor-sum arithmetic-functions multiplicative-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think brute search is optimal, if a bit tedious. You'll find the answer fairly quickly.
    $endgroup$
    – lulu
    Dec 9 '18 at 13:35










  • $begingroup$
    Oh, thanks, my method sounds good though.
    $endgroup$
    – Maged Saeed
    Dec 9 '18 at 13:35






  • 1




    $begingroup$
    Absolutely, yes. And you are correct about $2,12$. The third part won't take you as long as you fear. Keep in mind $sigma_1(n)≥n+1$ with equality only for primes. That makes it easy to truncate your search.
    $endgroup$
    – lulu
    Dec 9 '18 at 13:36








  • 1




    $begingroup$
    Correct again. $,$
    $endgroup$
    – lulu
    Dec 9 '18 at 13:39






  • 1




    $begingroup$
    (+1) for the posted solution, good work.
    $endgroup$
    – lulu
    Dec 9 '18 at 13:49














3












3








3


1



$begingroup$


The question says:




Find the smallest positive integer $n$ so that $sigma(x)=n$ has no solution, exactly two solutions, exactly three solutions.




I could not come up with a good way to solve this question other that trial and error. But I am questioning this method. Is there any better ideas?






number-theory elementary-number-theory divisor-sum arithmetic-functions multiplicative-function







share|cite|improve this question











$endgroup$




The question says:




Find the smallest positive integer $n$ so that $sigma(x)=n$ has no solution, exactly two solutions, exactly three solutions.




I could not come up with a good way to solve this question other that trial and error. But I am questioning this method. Is there any better ideas?






number-theory elementary-number-theory divisor-sum arithmetic-functions multiplicative-function




number-theory elementary-number-theory divisor-sum arithmetic-functions multiplicative-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 13:44







Maged Saeed

















asked Dec 9 '18 at 13:23









Maged SaeedMaged Saeed

8471417




8471417








  • 1




    $begingroup$
    I think brute search is optimal, if a bit tedious. You'll find the answer fairly quickly.
    $endgroup$
    – lulu
    Dec 9 '18 at 13:35










  • $begingroup$
    Oh, thanks, my method sounds good though.
    $endgroup$
    – Maged Saeed
    Dec 9 '18 at 13:35






  • 1




    $begingroup$
    Absolutely, yes. And you are correct about $2,12$. The third part won't take you as long as you fear. Keep in mind $sigma_1(n)≥n+1$ with equality only for primes. That makes it easy to truncate your search.
    $endgroup$
    – lulu
    Dec 9 '18 at 13:36








  • 1




    $begingroup$
    Correct again. $,$
    $endgroup$
    – lulu
    Dec 9 '18 at 13:39






  • 1




    $begingroup$
    (+1) for the posted solution, good work.
    $endgroup$
    – lulu
    Dec 9 '18 at 13:49














  • 1




    $begingroup$
    I think brute search is optimal, if a bit tedious. You'll find the answer fairly quickly.
    $endgroup$
    – lulu
    Dec 9 '18 at 13:35










  • $begingroup$
    Oh, thanks, my method sounds good though.
    $endgroup$
    – Maged Saeed
    Dec 9 '18 at 13:35






  • 1




    $begingroup$
    Absolutely, yes. And you are correct about $2,12$. The third part won't take you as long as you fear. Keep in mind $sigma_1(n)≥n+1$ with equality only for primes. That makes it easy to truncate your search.
    $endgroup$
    – lulu
    Dec 9 '18 at 13:36








  • 1




    $begingroup$
    Correct again. $,$
    $endgroup$
    – lulu
    Dec 9 '18 at 13:39






  • 1




    $begingroup$
    (+1) for the posted solution, good work.
    $endgroup$
    – lulu
    Dec 9 '18 at 13:49








1




1




$begingroup$
I think brute search is optimal, if a bit tedious. You'll find the answer fairly quickly.
$endgroup$
– lulu
Dec 9 '18 at 13:35




$begingroup$
I think brute search is optimal, if a bit tedious. You'll find the answer fairly quickly.
$endgroup$
– lulu
Dec 9 '18 at 13:35












$begingroup$
Oh, thanks, my method sounds good though.
$endgroup$
– Maged Saeed
Dec 9 '18 at 13:35




$begingroup$
Oh, thanks, my method sounds good though.
$endgroup$
– Maged Saeed
Dec 9 '18 at 13:35




1




1




$begingroup$
Absolutely, yes. And you are correct about $2,12$. The third part won't take you as long as you fear. Keep in mind $sigma_1(n)≥n+1$ with equality only for primes. That makes it easy to truncate your search.
$endgroup$
– lulu
Dec 9 '18 at 13:36






$begingroup$
Absolutely, yes. And you are correct about $2,12$. The third part won't take you as long as you fear. Keep in mind $sigma_1(n)≥n+1$ with equality only for primes. That makes it easy to truncate your search.
$endgroup$
– lulu
Dec 9 '18 at 13:36






1




1




$begingroup$
Correct again. $,$
$endgroup$
– lulu
Dec 9 '18 at 13:39




$begingroup$
Correct again. $,$
$endgroup$
– lulu
Dec 9 '18 at 13:39




1




1




$begingroup$
(+1) for the posted solution, good work.
$endgroup$
– lulu
Dec 9 '18 at 13:49




$begingroup$
(+1) for the posted solution, good work.
$endgroup$
– lulu
Dec 9 '18 at 13:49










2 Answers
2






active

oldest

votes


















4












$begingroup$

By trial and error, I have found that the solutions are:





  • $sigma(x) = 2$ has no solution.


  • $sigma(x) = 12$ has exactly two solutions that are $6$ and $11$.


  • $sigma(x) = 24$ has exactly three solutions that are $14,15$ and $23$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The number of divisors of a natural number $n$ is given by $sigma(n) = sum_{k=1}^{infty}left ( left lfloor frac{n}{k} right rfloor-left lfloor frac{n-1}{k} right rfloor right )$.



    This may be useful when expanding the summation.



    Note that, when $n$ is a prime number, $sum_{k=1}^{infty}left ( left lfloor frac{n}{k} right rfloor-left lfloor frac{n-1}{k} right rfloor right )=2$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      how did you come up with this formula?
      $endgroup$
      – Maged Saeed
      Dec 12 '18 at 2:03











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    2 Answers
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    2 Answers
    2






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    oldest

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    active

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    4












    $begingroup$

    By trial and error, I have found that the solutions are:





    • $sigma(x) = 2$ has no solution.


    • $sigma(x) = 12$ has exactly two solutions that are $6$ and $11$.


    • $sigma(x) = 24$ has exactly three solutions that are $14,15$ and $23$.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      By trial and error, I have found that the solutions are:





      • $sigma(x) = 2$ has no solution.


      • $sigma(x) = 12$ has exactly two solutions that are $6$ and $11$.


      • $sigma(x) = 24$ has exactly three solutions that are $14,15$ and $23$.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        By trial and error, I have found that the solutions are:





        • $sigma(x) = 2$ has no solution.


        • $sigma(x) = 12$ has exactly two solutions that are $6$ and $11$.


        • $sigma(x) = 24$ has exactly three solutions that are $14,15$ and $23$.






        share|cite|improve this answer









        $endgroup$



        By trial and error, I have found that the solutions are:





        • $sigma(x) = 2$ has no solution.


        • $sigma(x) = 12$ has exactly two solutions that are $6$ and $11$.


        • $sigma(x) = 24$ has exactly three solutions that are $14,15$ and $23$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 13:46









        Maged SaeedMaged Saeed

        8471417




        8471417























            1












            $begingroup$

            The number of divisors of a natural number $n$ is given by $sigma(n) = sum_{k=1}^{infty}left ( left lfloor frac{n}{k} right rfloor-left lfloor frac{n-1}{k} right rfloor right )$.



            This may be useful when expanding the summation.



            Note that, when $n$ is a prime number, $sum_{k=1}^{infty}left ( left lfloor frac{n}{k} right rfloor-left lfloor frac{n-1}{k} right rfloor right )=2$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              how did you come up with this formula?
              $endgroup$
              – Maged Saeed
              Dec 12 '18 at 2:03
















            1












            $begingroup$

            The number of divisors of a natural number $n$ is given by $sigma(n) = sum_{k=1}^{infty}left ( left lfloor frac{n}{k} right rfloor-left lfloor frac{n-1}{k} right rfloor right )$.



            This may be useful when expanding the summation.



            Note that, when $n$ is a prime number, $sum_{k=1}^{infty}left ( left lfloor frac{n}{k} right rfloor-left lfloor frac{n-1}{k} right rfloor right )=2$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              how did you come up with this formula?
              $endgroup$
              – Maged Saeed
              Dec 12 '18 at 2:03














            1












            1








            1





            $begingroup$

            The number of divisors of a natural number $n$ is given by $sigma(n) = sum_{k=1}^{infty}left ( left lfloor frac{n}{k} right rfloor-left lfloor frac{n-1}{k} right rfloor right )$.



            This may be useful when expanding the summation.



            Note that, when $n$ is a prime number, $sum_{k=1}^{infty}left ( left lfloor frac{n}{k} right rfloor-left lfloor frac{n-1}{k} right rfloor right )=2$






            share|cite|improve this answer









            $endgroup$



            The number of divisors of a natural number $n$ is given by $sigma(n) = sum_{k=1}^{infty}left ( left lfloor frac{n}{k} right rfloor-left lfloor frac{n-1}{k} right rfloor right )$.



            This may be useful when expanding the summation.



            Note that, when $n$ is a prime number, $sum_{k=1}^{infty}left ( left lfloor frac{n}{k} right rfloor-left lfloor frac{n-1}{k} right rfloor right )=2$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 9 '18 at 13:53









            Hussain-AlqatariHussain-Alqatari

            3187




            3187












            • $begingroup$
              how did you come up with this formula?
              $endgroup$
              – Maged Saeed
              Dec 12 '18 at 2:03


















            • $begingroup$
              how did you come up with this formula?
              $endgroup$
              – Maged Saeed
              Dec 12 '18 at 2:03
















            $begingroup$
            how did you come up with this formula?
            $endgroup$
            – Maged Saeed
            Dec 12 '18 at 2:03




            $begingroup$
            how did you come up with this formula?
            $endgroup$
            – Maged Saeed
            Dec 12 '18 at 2:03


















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