Number of divisors of $10!$












2












$begingroup$



Determine the amount of divisors of $10!$




This is a question in my combinatorics textbook, so I need to somehow reduce this to an elementary counting problem like combinations, permutations with or without repetition. I just don't see it. I can determine $10$ easy divisors, those are all the numbers $1$ through $10$ themselves. Then we can consider all possible products of these numbers, but then I get that a number can be either in the product , or not. I would get that it is $2^{10}=1024$ but the answer says it should be $270$.



What is going wrong here?



Note: $1$ and $10!$ are included






combinatorics divisibility factorial







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    You just have to combine $$ d(n)=prod_{pmid n}left(nu_p(n)+1right) $$ with $$ nu_p(n!)=sum_{kgeq 1}leftlfloor frac{n}{p^k}rightrfloor$$ to get that the answer is $$ (5+2+1+1)(3+1+1)(2+1)(1+1)=270.$$
    $endgroup$
    – Jack D'Aurizio
    Dec 11 '18 at 0:53


















2












$begingroup$



Determine the amount of divisors of $10!$




This is a question in my combinatorics textbook, so I need to somehow reduce this to an elementary counting problem like combinations, permutations with or without repetition. I just don't see it. I can determine $10$ easy divisors, those are all the numbers $1$ through $10$ themselves. Then we can consider all possible products of these numbers, but then I get that a number can be either in the product , or not. I would get that it is $2^{10}=1024$ but the answer says it should be $270$.



What is going wrong here?



Note: $1$ and $10!$ are included






combinatorics divisibility factorial







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    You just have to combine $$ d(n)=prod_{pmid n}left(nu_p(n)+1right) $$ with $$ nu_p(n!)=sum_{kgeq 1}leftlfloor frac{n}{p^k}rightrfloor$$ to get that the answer is $$ (5+2+1+1)(3+1+1)(2+1)(1+1)=270.$$
    $endgroup$
    – Jack D'Aurizio
    Dec 11 '18 at 0:53
















2












2








2





$begingroup$



Determine the amount of divisors of $10!$




This is a question in my combinatorics textbook, so I need to somehow reduce this to an elementary counting problem like combinations, permutations with or without repetition. I just don't see it. I can determine $10$ easy divisors, those are all the numbers $1$ through $10$ themselves. Then we can consider all possible products of these numbers, but then I get that a number can be either in the product , or not. I would get that it is $2^{10}=1024$ but the answer says it should be $270$.



What is going wrong here?



Note: $1$ and $10!$ are included






combinatorics divisibility factorial







share|cite|improve this question









$endgroup$





Determine the amount of divisors of $10!$




This is a question in my combinatorics textbook, so I need to somehow reduce this to an elementary counting problem like combinations, permutations with or without repetition. I just don't see it. I can determine $10$ easy divisors, those are all the numbers $1$ through $10$ themselves. Then we can consider all possible products of these numbers, but then I get that a number can be either in the product , or not. I would get that it is $2^{10}=1024$ but the answer says it should be $270$.



What is going wrong here?



Note: $1$ and $10!$ are included






combinatorics divisibility factorial




combinatorics divisibility factorial






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 11 '18 at 0:11









Wesley StrikWesley Strik

2,194424




2,194424








  • 2




    $begingroup$
    You just have to combine $$ d(n)=prod_{pmid n}left(nu_p(n)+1right) $$ with $$ nu_p(n!)=sum_{kgeq 1}leftlfloor frac{n}{p^k}rightrfloor$$ to get that the answer is $$ (5+2+1+1)(3+1+1)(2+1)(1+1)=270.$$
    $endgroup$
    – Jack D'Aurizio
    Dec 11 '18 at 0:53
















  • 2




    $begingroup$
    You just have to combine $$ d(n)=prod_{pmid n}left(nu_p(n)+1right) $$ with $$ nu_p(n!)=sum_{kgeq 1}leftlfloor frac{n}{p^k}rightrfloor$$ to get that the answer is $$ (5+2+1+1)(3+1+1)(2+1)(1+1)=270.$$
    $endgroup$
    – Jack D'Aurizio
    Dec 11 '18 at 0:53










2




2




$begingroup$
You just have to combine $$ d(n)=prod_{pmid n}left(nu_p(n)+1right) $$ with $$ nu_p(n!)=sum_{kgeq 1}leftlfloor frac{n}{p^k}rightrfloor$$ to get that the answer is $$ (5+2+1+1)(3+1+1)(2+1)(1+1)=270.$$
$endgroup$
– Jack D'Aurizio
Dec 11 '18 at 0:53






$begingroup$
You just have to combine $$ d(n)=prod_{pmid n}left(nu_p(n)+1right) $$ with $$ nu_p(n!)=sum_{kgeq 1}leftlfloor frac{n}{p^k}rightrfloor$$ to get that the answer is $$ (5+2+1+1)(3+1+1)(2+1)(1+1)=270.$$
$endgroup$
– Jack D'Aurizio
Dec 11 '18 at 0:53












3 Answers
3






active

oldest

votes


















7












$begingroup$

You've overcounted quite a bit, because integer factorization is not unique unless we're talking about primes. So for example, you've got $4 cdot 10 = 40 = 5 cdot 8$ counted twice.




So rather than thinking about numbers between $1$ and $10$, think of prime powers. A number is a divisor of $10!$ if and only if it is of the form



$$n = 2^a cdot 3^b cdot 5^c cdot 7^d$$
for appropriate ranges of $a, b, c, $ and $d$. I'll leave it to you to figure out why the values of $a, b, c, $ and $d$ range from $0$ to $8$, $4$, $2$ and $1$ respectively, giving



$$(8 + 1)(4 + 1)(2 + 1)(1 + 1) = 270$$



in total.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Easy, after your hint it all became clear, thank you! $10! = 1 cdot 2 cdot3 cdot 4 cdot 5 cdot 6 cdot 7 cdot 8cdot 9 cdot 10= 2 cdot 3 cdot 2^2 cdot 5 cdot 2 cdot 3 cdot 7 cdot 2^3 cdot 3^2 cdot 5 cdot 2= 2^8 cdot 3^4 cdot 5^2 cdot 7$
    $endgroup$
    – Wesley Strik
    Dec 11 '18 at 0:23








  • 1




    $begingroup$
    You're very welcome.
    $endgroup$
    – T. Bongers
    Dec 11 '18 at 0:24



















3












$begingroup$

Hint
$$10!=2^{??}cdot 3^{??}cdot5^{??}cdot 7^{??}$$



Now, any divisor must have the same primes with different powers...



The issue with your approach is that you are double and triple counting some divisors.



For example, you counted $8$ as $8$ but then you also counted it as $2 cdot 4$. Same way, most numbers divisible by $8$ are at least double counted.



You counted $24$ as the products $3 cdot 8, 4 cdot 6, 2 cdot 3 cdot 4$ and so on...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Of course, unique prime factorisation.
    $endgroup$
    – Wesley Strik
    Dec 11 '18 at 0:20



















0












$begingroup$

Hint for a smaller number: consider $2^3cdot 3^2$ = 72. To make one of its factors, how many times do you want to use $2$?






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    You've overcounted quite a bit, because integer factorization is not unique unless we're talking about primes. So for example, you've got $4 cdot 10 = 40 = 5 cdot 8$ counted twice.




    So rather than thinking about numbers between $1$ and $10$, think of prime powers. A number is a divisor of $10!$ if and only if it is of the form



    $$n = 2^a cdot 3^b cdot 5^c cdot 7^d$$
    for appropriate ranges of $a, b, c, $ and $d$. I'll leave it to you to figure out why the values of $a, b, c, $ and $d$ range from $0$ to $8$, $4$, $2$ and $1$ respectively, giving



    $$(8 + 1)(4 + 1)(2 + 1)(1 + 1) = 270$$



    in total.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Easy, after your hint it all became clear, thank you! $10! = 1 cdot 2 cdot3 cdot 4 cdot 5 cdot 6 cdot 7 cdot 8cdot 9 cdot 10= 2 cdot 3 cdot 2^2 cdot 5 cdot 2 cdot 3 cdot 7 cdot 2^3 cdot 3^2 cdot 5 cdot 2= 2^8 cdot 3^4 cdot 5^2 cdot 7$
      $endgroup$
      – Wesley Strik
      Dec 11 '18 at 0:23








    • 1




      $begingroup$
      You're very welcome.
      $endgroup$
      – T. Bongers
      Dec 11 '18 at 0:24
















    7












    $begingroup$

    You've overcounted quite a bit, because integer factorization is not unique unless we're talking about primes. So for example, you've got $4 cdot 10 = 40 = 5 cdot 8$ counted twice.




    So rather than thinking about numbers between $1$ and $10$, think of prime powers. A number is a divisor of $10!$ if and only if it is of the form



    $$n = 2^a cdot 3^b cdot 5^c cdot 7^d$$
    for appropriate ranges of $a, b, c, $ and $d$. I'll leave it to you to figure out why the values of $a, b, c, $ and $d$ range from $0$ to $8$, $4$, $2$ and $1$ respectively, giving



    $$(8 + 1)(4 + 1)(2 + 1)(1 + 1) = 270$$



    in total.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Easy, after your hint it all became clear, thank you! $10! = 1 cdot 2 cdot3 cdot 4 cdot 5 cdot 6 cdot 7 cdot 8cdot 9 cdot 10= 2 cdot 3 cdot 2^2 cdot 5 cdot 2 cdot 3 cdot 7 cdot 2^3 cdot 3^2 cdot 5 cdot 2= 2^8 cdot 3^4 cdot 5^2 cdot 7$
      $endgroup$
      – Wesley Strik
      Dec 11 '18 at 0:23








    • 1




      $begingroup$
      You're very welcome.
      $endgroup$
      – T. Bongers
      Dec 11 '18 at 0:24














    7












    7








    7





    $begingroup$

    You've overcounted quite a bit, because integer factorization is not unique unless we're talking about primes. So for example, you've got $4 cdot 10 = 40 = 5 cdot 8$ counted twice.




    So rather than thinking about numbers between $1$ and $10$, think of prime powers. A number is a divisor of $10!$ if and only if it is of the form



    $$n = 2^a cdot 3^b cdot 5^c cdot 7^d$$
    for appropriate ranges of $a, b, c, $ and $d$. I'll leave it to you to figure out why the values of $a, b, c, $ and $d$ range from $0$ to $8$, $4$, $2$ and $1$ respectively, giving



    $$(8 + 1)(4 + 1)(2 + 1)(1 + 1) = 270$$



    in total.






    share|cite|improve this answer









    $endgroup$



    You've overcounted quite a bit, because integer factorization is not unique unless we're talking about primes. So for example, you've got $4 cdot 10 = 40 = 5 cdot 8$ counted twice.




    So rather than thinking about numbers between $1$ and $10$, think of prime powers. A number is a divisor of $10!$ if and only if it is of the form



    $$n = 2^a cdot 3^b cdot 5^c cdot 7^d$$
    for appropriate ranges of $a, b, c, $ and $d$. I'll leave it to you to figure out why the values of $a, b, c, $ and $d$ range from $0$ to $8$, $4$, $2$ and $1$ respectively, giving



    $$(8 + 1)(4 + 1)(2 + 1)(1 + 1) = 270$$



    in total.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 11 '18 at 0:15









    T. BongersT. Bongers

    23.5k54762




    23.5k54762












    • $begingroup$
      Easy, after your hint it all became clear, thank you! $10! = 1 cdot 2 cdot3 cdot 4 cdot 5 cdot 6 cdot 7 cdot 8cdot 9 cdot 10= 2 cdot 3 cdot 2^2 cdot 5 cdot 2 cdot 3 cdot 7 cdot 2^3 cdot 3^2 cdot 5 cdot 2= 2^8 cdot 3^4 cdot 5^2 cdot 7$
      $endgroup$
      – Wesley Strik
      Dec 11 '18 at 0:23








    • 1




      $begingroup$
      You're very welcome.
      $endgroup$
      – T. Bongers
      Dec 11 '18 at 0:24


















    • $begingroup$
      Easy, after your hint it all became clear, thank you! $10! = 1 cdot 2 cdot3 cdot 4 cdot 5 cdot 6 cdot 7 cdot 8cdot 9 cdot 10= 2 cdot 3 cdot 2^2 cdot 5 cdot 2 cdot 3 cdot 7 cdot 2^3 cdot 3^2 cdot 5 cdot 2= 2^8 cdot 3^4 cdot 5^2 cdot 7$
      $endgroup$
      – Wesley Strik
      Dec 11 '18 at 0:23








    • 1




      $begingroup$
      You're very welcome.
      $endgroup$
      – T. Bongers
      Dec 11 '18 at 0:24
















    $begingroup$
    Easy, after your hint it all became clear, thank you! $10! = 1 cdot 2 cdot3 cdot 4 cdot 5 cdot 6 cdot 7 cdot 8cdot 9 cdot 10= 2 cdot 3 cdot 2^2 cdot 5 cdot 2 cdot 3 cdot 7 cdot 2^3 cdot 3^2 cdot 5 cdot 2= 2^8 cdot 3^4 cdot 5^2 cdot 7$
    $endgroup$
    – Wesley Strik
    Dec 11 '18 at 0:23






    $begingroup$
    Easy, after your hint it all became clear, thank you! $10! = 1 cdot 2 cdot3 cdot 4 cdot 5 cdot 6 cdot 7 cdot 8cdot 9 cdot 10= 2 cdot 3 cdot 2^2 cdot 5 cdot 2 cdot 3 cdot 7 cdot 2^3 cdot 3^2 cdot 5 cdot 2= 2^8 cdot 3^4 cdot 5^2 cdot 7$
    $endgroup$
    – Wesley Strik
    Dec 11 '18 at 0:23






    1




    1




    $begingroup$
    You're very welcome.
    $endgroup$
    – T. Bongers
    Dec 11 '18 at 0:24




    $begingroup$
    You're very welcome.
    $endgroup$
    – T. Bongers
    Dec 11 '18 at 0:24











    3












    $begingroup$

    Hint
    $$10!=2^{??}cdot 3^{??}cdot5^{??}cdot 7^{??}$$



    Now, any divisor must have the same primes with different powers...



    The issue with your approach is that you are double and triple counting some divisors.



    For example, you counted $8$ as $8$ but then you also counted it as $2 cdot 4$. Same way, most numbers divisible by $8$ are at least double counted.



    You counted $24$ as the products $3 cdot 8, 4 cdot 6, 2 cdot 3 cdot 4$ and so on...






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Of course, unique prime factorisation.
      $endgroup$
      – Wesley Strik
      Dec 11 '18 at 0:20
















    3












    $begingroup$

    Hint
    $$10!=2^{??}cdot 3^{??}cdot5^{??}cdot 7^{??}$$



    Now, any divisor must have the same primes with different powers...



    The issue with your approach is that you are double and triple counting some divisors.



    For example, you counted $8$ as $8$ but then you also counted it as $2 cdot 4$. Same way, most numbers divisible by $8$ are at least double counted.



    You counted $24$ as the products $3 cdot 8, 4 cdot 6, 2 cdot 3 cdot 4$ and so on...






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Of course, unique prime factorisation.
      $endgroup$
      – Wesley Strik
      Dec 11 '18 at 0:20














    3












    3








    3





    $begingroup$

    Hint
    $$10!=2^{??}cdot 3^{??}cdot5^{??}cdot 7^{??}$$



    Now, any divisor must have the same primes with different powers...



    The issue with your approach is that you are double and triple counting some divisors.



    For example, you counted $8$ as $8$ but then you also counted it as $2 cdot 4$. Same way, most numbers divisible by $8$ are at least double counted.



    You counted $24$ as the products $3 cdot 8, 4 cdot 6, 2 cdot 3 cdot 4$ and so on...






    share|cite|improve this answer











    $endgroup$



    Hint
    $$10!=2^{??}cdot 3^{??}cdot5^{??}cdot 7^{??}$$



    Now, any divisor must have the same primes with different powers...



    The issue with your approach is that you are double and triple counting some divisors.



    For example, you counted $8$ as $8$ but then you also counted it as $2 cdot 4$. Same way, most numbers divisible by $8$ are at least double counted.



    You counted $24$ as the products $3 cdot 8, 4 cdot 6, 2 cdot 3 cdot 4$ and so on...







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 11 '18 at 1:25









    AryanSonwatikar

    471114




    471114










    answered Dec 11 '18 at 0:18









    N. S.N. S.

    105k7114210




    105k7114210












    • $begingroup$
      Of course, unique prime factorisation.
      $endgroup$
      – Wesley Strik
      Dec 11 '18 at 0:20


















    • $begingroup$
      Of course, unique prime factorisation.
      $endgroup$
      – Wesley Strik
      Dec 11 '18 at 0:20
















    $begingroup$
    Of course, unique prime factorisation.
    $endgroup$
    – Wesley Strik
    Dec 11 '18 at 0:20




    $begingroup$
    Of course, unique prime factorisation.
    $endgroup$
    – Wesley Strik
    Dec 11 '18 at 0:20











    0












    $begingroup$

    Hint for a smaller number: consider $2^3cdot 3^2$ = 72. To make one of its factors, how many times do you want to use $2$?






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint for a smaller number: consider $2^3cdot 3^2$ = 72. To make one of its factors, how many times do you want to use $2$?






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint for a smaller number: consider $2^3cdot 3^2$ = 72. To make one of its factors, how many times do you want to use $2$?






        share|cite|improve this answer









        $endgroup$



        Hint for a smaller number: consider $2^3cdot 3^2$ = 72. To make one of its factors, how many times do you want to use $2$?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 1:18









        timtfjtimtfj

        2,483420




        2,483420






























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