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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034686%2fnumber-of-divisors-of-10%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034686%2fnumber-of-divisors-of-10%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
            Read more