Finding the sum of 3 numbers with 5 positive integer divisors. The 2019 Stack Overflow...

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Finding the sum of 3 numbers with 5 positive integer divisors.



The 2019 Stack Overflow Developer Survey Results Are InNumber theory with positive integer $n$ questionUpper bound for the difference between number-of-divisors and sum-of-divisors functionsSmallest positive integer with 60 divisorsFind the least positive integer with $24$ positive divisors.Counting and bounding square-free numbers formed from only the first $j$ primesThe product of five consecutive positive integers cannot be the square of an integerA conjecture concerning the number of divisors and the sum of divisors.Range of the map “product of positive divisors”Which natural numbers are the sum of three positive perfect squares?About The Sum of Positive Divisors of $n$












1












$begingroup$



What is the sum of the three numbers less that 1000 that have exactly
five positive integer divisors?




So I see that first since there are only 5 multiples, the number must be a perfect square. Then, I took a look at a few examples.



$#1text{:}$



$$2^2=4$$



$$impliestext{1, 4}$$



$$impliestext{2, 2}$$



4 then has 3 positive integer divisors.



$#2text{:}$



$$4^2=16$$



$$impliestext{1, 16}$$



$$impliestext{2, 8}$$



$$impliestext{4, 4}$$



This means that the square of a square number has 5 positive integer divisors. There are many of these under 1000, so how come there are three? I am probably missing a point here. What is wrong with my logic? How can I solve this problem?






number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your last conclusion is not right. For instance, $256$ is a square of a square number, but has many more than five positive integer divisors. Hint: prime number factorization.
    $endgroup$
    – Brian Tung
    Mar 22 at 0:12










  • $begingroup$
    Incidentally, there aren't that many fourth powers less than $1000$, but there are more than three. So you do need to narrow it down.
    $endgroup$
    – Brian Tung
    Mar 22 at 0:14
















1












$begingroup$



What is the sum of the three numbers less that 1000 that have exactly
five positive integer divisors?




So I see that first since there are only 5 multiples, the number must be a perfect square. Then, I took a look at a few examples.



$#1text{:}$



$$2^2=4$$



$$impliestext{1, 4}$$



$$impliestext{2, 2}$$



4 then has 3 positive integer divisors.



$#2text{:}$



$$4^2=16$$



$$impliestext{1, 16}$$



$$impliestext{2, 8}$$



$$impliestext{4, 4}$$



This means that the square of a square number has 5 positive integer divisors. There are many of these under 1000, so how come there are three? I am probably missing a point here. What is wrong with my logic? How can I solve this problem?






number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your last conclusion is not right. For instance, $256$ is a square of a square number, but has many more than five positive integer divisors. Hint: prime number factorization.
    $endgroup$
    – Brian Tung
    Mar 22 at 0:12










  • $begingroup$
    Incidentally, there aren't that many fourth powers less than $1000$, but there are more than three. So you do need to narrow it down.
    $endgroup$
    – Brian Tung
    Mar 22 at 0:14














1












1








1





$begingroup$



What is the sum of the three numbers less that 1000 that have exactly
five positive integer divisors?




So I see that first since there are only 5 multiples, the number must be a perfect square. Then, I took a look at a few examples.



$#1text{:}$



$$2^2=4$$



$$impliestext{1, 4}$$



$$impliestext{2, 2}$$



4 then has 3 positive integer divisors.



$#2text{:}$



$$4^2=16$$



$$impliestext{1, 16}$$



$$impliestext{2, 8}$$



$$impliestext{4, 4}$$



This means that the square of a square number has 5 positive integer divisors. There are many of these under 1000, so how come there are three? I am probably missing a point here. What is wrong with my logic? How can I solve this problem?






number-theory







share|cite|improve this question









$endgroup$





What is the sum of the three numbers less that 1000 that have exactly
five positive integer divisors?




So I see that first since there are only 5 multiples, the number must be a perfect square. Then, I took a look at a few examples.



$#1text{:}$



$$2^2=4$$



$$impliestext{1, 4}$$



$$impliestext{2, 2}$$



4 then has 3 positive integer divisors.



$#2text{:}$



$$4^2=16$$



$$impliestext{1, 16}$$



$$impliestext{2, 8}$$



$$impliestext{4, 4}$$



This means that the square of a square number has 5 positive integer divisors. There are many of these under 1000, so how come there are three? I am probably missing a point here. What is wrong with my logic? How can I solve this problem?






number-theory




number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 22 at 0:08









Max0815Max0815

81418




81418












  • $begingroup$
    Your last conclusion is not right. For instance, $256$ is a square of a square number, but has many more than five positive integer divisors. Hint: prime number factorization.
    $endgroup$
    – Brian Tung
    Mar 22 at 0:12










  • $begingroup$
    Incidentally, there aren't that many fourth powers less than $1000$, but there are more than three. So you do need to narrow it down.
    $endgroup$
    – Brian Tung
    Mar 22 at 0:14


















  • $begingroup$
    Your last conclusion is not right. For instance, $256$ is a square of a square number, but has many more than five positive integer divisors. Hint: prime number factorization.
    $endgroup$
    – Brian Tung
    Mar 22 at 0:12










  • $begingroup$
    Incidentally, there aren't that many fourth powers less than $1000$, but there are more than three. So you do need to narrow it down.
    $endgroup$
    – Brian Tung
    Mar 22 at 0:14
















$begingroup$
Your last conclusion is not right. For instance, $256$ is a square of a square number, but has many more than five positive integer divisors. Hint: prime number factorization.
$endgroup$
– Brian Tung
Mar 22 at 0:12




$begingroup$
Your last conclusion is not right. For instance, $256$ is a square of a square number, but has many more than five positive integer divisors. Hint: prime number factorization.
$endgroup$
– Brian Tung
Mar 22 at 0:12












$begingroup$
Incidentally, there aren't that many fourth powers less than $1000$, but there are more than three. So you do need to narrow it down.
$endgroup$
– Brian Tung
Mar 22 at 0:14




$begingroup$
Incidentally, there aren't that many fourth powers less than $1000$, but there are more than three. So you do need to narrow it down.
$endgroup$
– Brian Tung
Mar 22 at 0:14










2 Answers
2






active

oldest

votes


















2












$begingroup$

Recall: The number of divisors of a number $n$ whose prime decomposition is



$$n=p_1^{alpha_1}cdots p_k^{alpha_k}$$



Is $(alpha_1+1)cdots(alpha_k+1)$ (Exercise: prove this statement), then, you want this number to be $5$ and each factor being positive you must have all to be one except for one of them that must be $5$.



$5$ is prime you can only express it as $1cdot 5$; so if every factor is one, that means the exponent of that prime is $0$ in the factorization of $n$, and the factor that is $5$ tells us there's only one prime in the factorization and that the power is $5-1=4$ (solving for $alpha_text{something}$).



With a little algebra you get that $n=p^4$ for some prime number $p$, then, the only prime numbers whose fourth power is less than $1000$ are $p=2,3,5$ so the sum of the four numbers is



$$2^4+3^4+5^4=722$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you explain how you got to that $n=p^4$?
    $endgroup$
    – Max0815
    Mar 22 at 0:54










  • $begingroup$
    @Max0815 Well, since $5$ is prime you can only express it as $1cdot5$; so if every factor is one, that means the exponent of that prime is $0$ in the factorization of $n$, and the factor that is $5$ tells us there's only one prime in the factorization and that the power is $5-1=4$ (solving for $alpha_{something}$)
    $endgroup$
    – Bruno Andrades
    Mar 22 at 0:58










  • $begingroup$
    thank you I see how you got that!
    $endgroup$
    – Max0815
    Mar 22 at 0:59










  • $begingroup$
    @Max0815 Glad to help! :)
    $endgroup$
    – Bruno Andrades
    Mar 22 at 0:59










  • $begingroup$
    I added the clarification part to the answer. Its useful to me. After the edit is accepeted, I'll accept your answer :)
    $endgroup$
    – Max0815
    Mar 22 at 1:02



















1












$begingroup$

The three numbers less than 1000 with exactly 5 factors are:




  • 16 {1, 2, 4, 8, 16}

  • 81 {1, 3, 9, 27, 81}

  • 625 {1, 5, 25, 125, 625}


So the sum is: 722, just as Max wrote.






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Recall: The number of divisors of a number $n$ whose prime decomposition is



    $$n=p_1^{alpha_1}cdots p_k^{alpha_k}$$



    Is $(alpha_1+1)cdots(alpha_k+1)$ (Exercise: prove this statement), then, you want this number to be $5$ and each factor being positive you must have all to be one except for one of them that must be $5$.



    $5$ is prime you can only express it as $1cdot 5$; so if every factor is one, that means the exponent of that prime is $0$ in the factorization of $n$, and the factor that is $5$ tells us there's only one prime in the factorization and that the power is $5-1=4$ (solving for $alpha_text{something}$).



    With a little algebra you get that $n=p^4$ for some prime number $p$, then, the only prime numbers whose fourth power is less than $1000$ are $p=2,3,5$ so the sum of the four numbers is



    $$2^4+3^4+5^4=722$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Can you explain how you got to that $n=p^4$?
      $endgroup$
      – Max0815
      Mar 22 at 0:54










    • $begingroup$
      @Max0815 Well, since $5$ is prime you can only express it as $1cdot5$; so if every factor is one, that means the exponent of that prime is $0$ in the factorization of $n$, and the factor that is $5$ tells us there's only one prime in the factorization and that the power is $5-1=4$ (solving for $alpha_{something}$)
      $endgroup$
      – Bruno Andrades
      Mar 22 at 0:58










    • $begingroup$
      thank you I see how you got that!
      $endgroup$
      – Max0815
      Mar 22 at 0:59










    • $begingroup$
      @Max0815 Glad to help! :)
      $endgroup$
      – Bruno Andrades
      Mar 22 at 0:59










    • $begingroup$
      I added the clarification part to the answer. Its useful to me. After the edit is accepeted, I'll accept your answer :)
      $endgroup$
      – Max0815
      Mar 22 at 1:02
















    2












    $begingroup$

    Recall: The number of divisors of a number $n$ whose prime decomposition is



    $$n=p_1^{alpha_1}cdots p_k^{alpha_k}$$



    Is $(alpha_1+1)cdots(alpha_k+1)$ (Exercise: prove this statement), then, you want this number to be $5$ and each factor being positive you must have all to be one except for one of them that must be $5$.



    $5$ is prime you can only express it as $1cdot 5$; so if every factor is one, that means the exponent of that prime is $0$ in the factorization of $n$, and the factor that is $5$ tells us there's only one prime in the factorization and that the power is $5-1=4$ (solving for $alpha_text{something}$).



    With a little algebra you get that $n=p^4$ for some prime number $p$, then, the only prime numbers whose fourth power is less than $1000$ are $p=2,3,5$ so the sum of the four numbers is



    $$2^4+3^4+5^4=722$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Can you explain how you got to that $n=p^4$?
      $endgroup$
      – Max0815
      Mar 22 at 0:54










    • $begingroup$
      @Max0815 Well, since $5$ is prime you can only express it as $1cdot5$; so if every factor is one, that means the exponent of that prime is $0$ in the factorization of $n$, and the factor that is $5$ tells us there's only one prime in the factorization and that the power is $5-1=4$ (solving for $alpha_{something}$)
      $endgroup$
      – Bruno Andrades
      Mar 22 at 0:58










    • $begingroup$
      thank you I see how you got that!
      $endgroup$
      – Max0815
      Mar 22 at 0:59










    • $begingroup$
      @Max0815 Glad to help! :)
      $endgroup$
      – Bruno Andrades
      Mar 22 at 0:59










    • $begingroup$
      I added the clarification part to the answer. Its useful to me. After the edit is accepeted, I'll accept your answer :)
      $endgroup$
      – Max0815
      Mar 22 at 1:02














    2












    2








    2





    $begingroup$

    Recall: The number of divisors of a number $n$ whose prime decomposition is



    $$n=p_1^{alpha_1}cdots p_k^{alpha_k}$$



    Is $(alpha_1+1)cdots(alpha_k+1)$ (Exercise: prove this statement), then, you want this number to be $5$ and each factor being positive you must have all to be one except for one of them that must be $5$.



    $5$ is prime you can only express it as $1cdot 5$; so if every factor is one, that means the exponent of that prime is $0$ in the factorization of $n$, and the factor that is $5$ tells us there's only one prime in the factorization and that the power is $5-1=4$ (solving for $alpha_text{something}$).



    With a little algebra you get that $n=p^4$ for some prime number $p$, then, the only prime numbers whose fourth power is less than $1000$ are $p=2,3,5$ so the sum of the four numbers is



    $$2^4+3^4+5^4=722$$






    share|cite|improve this answer











    $endgroup$



    Recall: The number of divisors of a number $n$ whose prime decomposition is



    $$n=p_1^{alpha_1}cdots p_k^{alpha_k}$$



    Is $(alpha_1+1)cdots(alpha_k+1)$ (Exercise: prove this statement), then, you want this number to be $5$ and each factor being positive you must have all to be one except for one of them that must be $5$.



    $5$ is prime you can only express it as $1cdot 5$; so if every factor is one, that means the exponent of that prime is $0$ in the factorization of $n$, and the factor that is $5$ tells us there's only one prime in the factorization and that the power is $5-1=4$ (solving for $alpha_text{something}$).



    With a little algebra you get that $n=p^4$ for some prime number $p$, then, the only prime numbers whose fourth power is less than $1000$ are $p=2,3,5$ so the sum of the four numbers is



    $$2^4+3^4+5^4=722$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 22 at 1:39









    Max0815

    81418




    81418










    answered Mar 22 at 0:24









    Bruno AndradesBruno Andrades

    24811




    24811












    • $begingroup$
      Can you explain how you got to that $n=p^4$?
      $endgroup$
      – Max0815
      Mar 22 at 0:54










    • $begingroup$
      @Max0815 Well, since $5$ is prime you can only express it as $1cdot5$; so if every factor is one, that means the exponent of that prime is $0$ in the factorization of $n$, and the factor that is $5$ tells us there's only one prime in the factorization and that the power is $5-1=4$ (solving for $alpha_{something}$)
      $endgroup$
      – Bruno Andrades
      Mar 22 at 0:58










    • $begingroup$
      thank you I see how you got that!
      $endgroup$
      – Max0815
      Mar 22 at 0:59










    • $begingroup$
      @Max0815 Glad to help! :)
      $endgroup$
      – Bruno Andrades
      Mar 22 at 0:59










    • $begingroup$
      I added the clarification part to the answer. Its useful to me. After the edit is accepeted, I'll accept your answer :)
      $endgroup$
      – Max0815
      Mar 22 at 1:02


















    • $begingroup$
      Can you explain how you got to that $n=p^4$?
      $endgroup$
      – Max0815
      Mar 22 at 0:54










    • $begingroup$
      @Max0815 Well, since $5$ is prime you can only express it as $1cdot5$; so if every factor is one, that means the exponent of that prime is $0$ in the factorization of $n$, and the factor that is $5$ tells us there's only one prime in the factorization and that the power is $5-1=4$ (solving for $alpha_{something}$)
      $endgroup$
      – Bruno Andrades
      Mar 22 at 0:58










    • $begingroup$
      thank you I see how you got that!
      $endgroup$
      – Max0815
      Mar 22 at 0:59










    • $begingroup$
      @Max0815 Glad to help! :)
      $endgroup$
      – Bruno Andrades
      Mar 22 at 0:59










    • $begingroup$
      I added the clarification part to the answer. Its useful to me. After the edit is accepeted, I'll accept your answer :)
      $endgroup$
      – Max0815
      Mar 22 at 1:02
















    $begingroup$
    Can you explain how you got to that $n=p^4$?
    $endgroup$
    – Max0815
    Mar 22 at 0:54




    $begingroup$
    Can you explain how you got to that $n=p^4$?
    $endgroup$
    – Max0815
    Mar 22 at 0:54












    $begingroup$
    @Max0815 Well, since $5$ is prime you can only express it as $1cdot5$; so if every factor is one, that means the exponent of that prime is $0$ in the factorization of $n$, and the factor that is $5$ tells us there's only one prime in the factorization and that the power is $5-1=4$ (solving for $alpha_{something}$)
    $endgroup$
    – Bruno Andrades
    Mar 22 at 0:58




    $begingroup$
    @Max0815 Well, since $5$ is prime you can only express it as $1cdot5$; so if every factor is one, that means the exponent of that prime is $0$ in the factorization of $n$, and the factor that is $5$ tells us there's only one prime in the factorization and that the power is $5-1=4$ (solving for $alpha_{something}$)
    $endgroup$
    – Bruno Andrades
    Mar 22 at 0:58












    $begingroup$
    thank you I see how you got that!
    $endgroup$
    – Max0815
    Mar 22 at 0:59




    $begingroup$
    thank you I see how you got that!
    $endgroup$
    – Max0815
    Mar 22 at 0:59












    $begingroup$
    @Max0815 Glad to help! :)
    $endgroup$
    – Bruno Andrades
    Mar 22 at 0:59




    $begingroup$
    @Max0815 Glad to help! :)
    $endgroup$
    – Bruno Andrades
    Mar 22 at 0:59












    $begingroup$
    I added the clarification part to the answer. Its useful to me. After the edit is accepeted, I'll accept your answer :)
    $endgroup$
    – Max0815
    Mar 22 at 1:02




    $begingroup$
    I added the clarification part to the answer. Its useful to me. After the edit is accepeted, I'll accept your answer :)
    $endgroup$
    – Max0815
    Mar 22 at 1:02











    1












    $begingroup$

    The three numbers less than 1000 with exactly 5 factors are:




    • 16 {1, 2, 4, 8, 16}

    • 81 {1, 3, 9, 27, 81}

    • 625 {1, 5, 25, 125, 625}


    So the sum is: 722, just as Max wrote.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The three numbers less than 1000 with exactly 5 factors are:




      • 16 {1, 2, 4, 8, 16}

      • 81 {1, 3, 9, 27, 81}

      • 625 {1, 5, 25, 125, 625}


      So the sum is: 722, just as Max wrote.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The three numbers less than 1000 with exactly 5 factors are:




        • 16 {1, 2, 4, 8, 16}

        • 81 {1, 3, 9, 27, 81}

        • 625 {1, 5, 25, 125, 625}


        So the sum is: 722, just as Max wrote.






        share|cite|improve this answer









        $endgroup$



        The three numbers less than 1000 with exactly 5 factors are:




        • 16 {1, 2, 4, 8, 16}

        • 81 {1, 3, 9, 27, 81}

        • 625 {1, 5, 25, 125, 625}


        So the sum is: 722, just as Max wrote.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 22 at 1:47









        David G. StorkDavid G. Stork

        12.1k41836




        12.1k41836






























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            Nidaros erkebispedøme

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            For det lutherske bispedømet, sjå Nidaros bispedøme. Nidaros erkebispedøme var eit katolsk erkebispedøme med sete i Nidaros. Det vart oppretta av kong Olav Kyrre mellom 1070 og 1080 som bispedøme. Bispedømet vart opphøgd til erkebispedøme i 1153. Erkebispedømet vart innstifta av pave Anastasius IV i 1154, og var i funksjon fram til reformasjonen i 1537. Erkebiskopen sitt hovudsete var Erkebispegarden, rett ved domkyrkja Nidarosdomen. Frå 1152 til 1154 var Nicholas Breakspear i Skandinavia som paveleg utsending, og organiserte opprettinga av erkesetet. Innhaldsliste 1 Utbreiing og jurisdiksjon 2 Overhyrdar 2.1 Lydbiskopar 3 Biskopar 3.1 Biskopar 3.2 Erkebiskopar Utbreiing og jurisdiksjon | Kart over den nordiske kyrkjeprovinsen (1153-1387) Erkebispedømet omfatta i tillegg til Noreg øyane i vest i Noregsveldet. Nidaros var eit metropolitanerkebispedøme med fylgjande underliggjande bispedøme suffraganar: Bjørgvin bispedøme Oslo bispedøme Hamar bispedøme Stavanger bispedøme Kirkjub
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            Birsay

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            Tettstaden Birsay på Orknøyane. Ruinane av jarleplasset om lag midt i biletet. (Fotografi frå 2012) Utsikt mot Birsay og Brough of Brisay (fotografi frå 2009) Birsay er ein tettstad i eit sokn (parish) med same namn, nordvest på øya Mainland [Note 1] på Orknøyane, om lag 31 km frå Kirkwall. Staden er omtala fleire stader i Orknøyingasoga [1] , der han er kalla «Byrgesherad» (norrønt: Birgisherað ). Namnet skriv seg truleg frå elvenamnet Birgisá [2] . Innhaldsliste 1 Den gamle hovudstaden 2 Bispesetet 3 Jarlepalasset 4 Kjelder og utfyllande opplysningar 4.1 Notar 4.2 Fotnotar 4.3 Bakgrunnsstoff Den gamle hovudstaden | Hovudartikkelen for dette emnet er Torfinn Sigurdsson . Soknet er i dag i hovudsak ei jordbrukskommune men er mest kjend for dei mange kulturminna i området. Tettstaden Birsay blir gjerne omtala som «den gamle hovudstaden på Orknøyane» [3] , og ein viser då gjerne til Orknøyingasoga (kap.31) som fortel at orknøyjarlen Torfinn Sigurdsson den mektige «hadde jamleg
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