Find the cardinality of set ${(x,y,z):x,y,z in mathbb{Z}^+ wedge xyz=108}$












0












$begingroup$


Find the cardinality of set ${(x,y,z): x,y,ztext{ are positive integers and }xyz=108}$



my attempt:



I write all factors of $108$.



$$1,2,3,4,6,9,12,18,27,36,54,108.$$
First observe that $xyz $ has only two degree of freedom. So its enough to choose $x,y$ . If I choose $x=1$ there are $12$ ways of choosing $y$. If choose $x$ as $2$ there are $11$ ways ...and so on.



So total number of ways $=1+2+...12=78$.
But its give in my book that $60$ is answer. Where am I wrong?






combinatorics permutations combinations integers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    "If I choose $2$ there are $11$ ways of choosing $y$" Really? I only found $8$ ways: $1,2,3,6,9,18,27,54$. What are the other $3$ ways???
    $endgroup$
    – bof
    Jan 15 at 7:36










  • $begingroup$
    @bof oh...thanks
    $endgroup$
    – Cloud JR
    Jan 15 at 7:38
















0












$begingroup$


Find the cardinality of set ${(x,y,z): x,y,ztext{ are positive integers and }xyz=108}$



my attempt:



I write all factors of $108$.



$$1,2,3,4,6,9,12,18,27,36,54,108.$$
First observe that $xyz $ has only two degree of freedom. So its enough to choose $x,y$ . If I choose $x=1$ there are $12$ ways of choosing $y$. If choose $x$ as $2$ there are $11$ ways ...and so on.



So total number of ways $=1+2+...12=78$.
But its give in my book that $60$ is answer. Where am I wrong?






combinatorics permutations combinations integers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    "If I choose $2$ there are $11$ ways of choosing $y$" Really? I only found $8$ ways: $1,2,3,6,9,18,27,54$. What are the other $3$ ways???
    $endgroup$
    – bof
    Jan 15 at 7:36










  • $begingroup$
    @bof oh...thanks
    $endgroup$
    – Cloud JR
    Jan 15 at 7:38














0












0








0





$begingroup$


Find the cardinality of set ${(x,y,z): x,y,ztext{ are positive integers and }xyz=108}$



my attempt:



I write all factors of $108$.



$$1,2,3,4,6,9,12,18,27,36,54,108.$$
First observe that $xyz $ has only two degree of freedom. So its enough to choose $x,y$ . If I choose $x=1$ there are $12$ ways of choosing $y$. If choose $x$ as $2$ there are $11$ ways ...and so on.



So total number of ways $=1+2+...12=78$.
But its give in my book that $60$ is answer. Where am I wrong?






combinatorics permutations combinations integers







share|cite|improve this question











$endgroup$




Find the cardinality of set ${(x,y,z): x,y,ztext{ are positive integers and }xyz=108}$



my attempt:



I write all factors of $108$.



$$1,2,3,4,6,9,12,18,27,36,54,108.$$
First observe that $xyz $ has only two degree of freedom. So its enough to choose $x,y$ . If I choose $x=1$ there are $12$ ways of choosing $y$. If choose $x$ as $2$ there are $11$ ways ...and so on.



So total number of ways $=1+2+...12=78$.
But its give in my book that $60$ is answer. Where am I wrong?






combinatorics permutations combinations integers




combinatorics permutations combinations integers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 7:52









David G. Stork

11k41432




11k41432










asked Jan 15 at 6:59









Cloud JRCloud JR

909518




909518








  • 1




    $begingroup$
    "If I choose $2$ there are $11$ ways of choosing $y$" Really? I only found $8$ ways: $1,2,3,6,9,18,27,54$. What are the other $3$ ways???
    $endgroup$
    – bof
    Jan 15 at 7:36










  • $begingroup$
    @bof oh...thanks
    $endgroup$
    – Cloud JR
    Jan 15 at 7:38














  • 1




    $begingroup$
    "If I choose $2$ there are $11$ ways of choosing $y$" Really? I only found $8$ ways: $1,2,3,6,9,18,27,54$. What are the other $3$ ways???
    $endgroup$
    – bof
    Jan 15 at 7:36










  • $begingroup$
    @bof oh...thanks
    $endgroup$
    – Cloud JR
    Jan 15 at 7:38








1




1




$begingroup$
"If I choose $2$ there are $11$ ways of choosing $y$" Really? I only found $8$ ways: $1,2,3,6,9,18,27,54$. What are the other $3$ ways???
$endgroup$
– bof
Jan 15 at 7:36




$begingroup$
"If I choose $2$ there are $11$ ways of choosing $y$" Really? I only found $8$ ways: $1,2,3,6,9,18,27,54$. What are the other $3$ ways???
$endgroup$
– bof
Jan 15 at 7:36












$begingroup$
@bof oh...thanks
$endgroup$
– Cloud JR
Jan 15 at 7:38




$begingroup$
@bof oh...thanks
$endgroup$
– Cloud JR
Jan 15 at 7:38










2 Answers
2






active

oldest

votes


















3












$begingroup$

$$(2^{j_1} 3^{k_1})(2^{j_2}3^{k_2})(2^{j_3}3^{k_3}) = 108 = 2^2 cdot 3^3$$



where ${ j_i, k_i} ge 0$ and by the Prime Factorization Theorem we have $j_1 + j_2 + j_3 = 2$ and $k_1 + k_2 + k_3 = 3$.



How many ways can you solve those equations, up to equivalent permutations of the factors? $12$



Start with the first (smallest) factor $(2^0 3^0)$ and increment others for the first "row"; then $(2^1 3^0)$ and increment others for the second "row"; then $(2^0 3^1)$ and so on... for the third "row." There is no fourth "row"...



Here are the sets, each in ascending order, themselves ordered:




  • {1, 1, 108}, {1, 2, 54}, {1, 3, 36}, {1, 4, 27}, {1, 6, 18}, {1, 9, 12}

  • {2, 2, 27}, {2, 3, 18}, {2, 6, 9}

  • {3, 3, 12}, {3, 4, 9}, {3, 6, 6}






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    $108=2^2 cdot 3^3$, so $x,y,z$ must be of the form $2^i3^j$ such that the sum of the $i$'s (where $0le ile 2$) is $2$ and the sum of the $j$'s (also integers starting from $0$) is $3$. How many solutions do $i_x+i_y+i_z = 2$ and $j_x+j_y+j_z = 3$ have in integers $ge 0$? All combinations of these solutions are the solutions you want.



    There are $binom{n+k-1}{k-1}$ non-negative integer solutions to $x_1+ ldots +x_k =n$ (see wikipedia, e.g.) and so for $3$ variables and $n=2$ we get $binom{2+3-1}{2} = binom{4}{2}=6$ and for $n=3$ we get $binom{5}{2} = 10$. So combined we get $6 times 10=60$ solutions for these independent equations.






    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%2f3074149%2ffind-the-cardinality-of-set-x-y-zx-y-z-in-mathbbz-wedge-xyz-108%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      $$(2^{j_1} 3^{k_1})(2^{j_2}3^{k_2})(2^{j_3}3^{k_3}) = 108 = 2^2 cdot 3^3$$



      where ${ j_i, k_i} ge 0$ and by the Prime Factorization Theorem we have $j_1 + j_2 + j_3 = 2$ and $k_1 + k_2 + k_3 = 3$.



      How many ways can you solve those equations, up to equivalent permutations of the factors? $12$



      Start with the first (smallest) factor $(2^0 3^0)$ and increment others for the first "row"; then $(2^1 3^0)$ and increment others for the second "row"; then $(2^0 3^1)$ and so on... for the third "row." There is no fourth "row"...



      Here are the sets, each in ascending order, themselves ordered:




      • {1, 1, 108}, {1, 2, 54}, {1, 3, 36}, {1, 4, 27}, {1, 6, 18}, {1, 9, 12}

      • {2, 2, 27}, {2, 3, 18}, {2, 6, 9}

      • {3, 3, 12}, {3, 4, 9}, {3, 6, 6}






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        $$(2^{j_1} 3^{k_1})(2^{j_2}3^{k_2})(2^{j_3}3^{k_3}) = 108 = 2^2 cdot 3^3$$



        where ${ j_i, k_i} ge 0$ and by the Prime Factorization Theorem we have $j_1 + j_2 + j_3 = 2$ and $k_1 + k_2 + k_3 = 3$.



        How many ways can you solve those equations, up to equivalent permutations of the factors? $12$



        Start with the first (smallest) factor $(2^0 3^0)$ and increment others for the first "row"; then $(2^1 3^0)$ and increment others for the second "row"; then $(2^0 3^1)$ and so on... for the third "row." There is no fourth "row"...



        Here are the sets, each in ascending order, themselves ordered:




        • {1, 1, 108}, {1, 2, 54}, {1, 3, 36}, {1, 4, 27}, {1, 6, 18}, {1, 9, 12}

        • {2, 2, 27}, {2, 3, 18}, {2, 6, 9}

        • {3, 3, 12}, {3, 4, 9}, {3, 6, 6}






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          $$(2^{j_1} 3^{k_1})(2^{j_2}3^{k_2})(2^{j_3}3^{k_3}) = 108 = 2^2 cdot 3^3$$



          where ${ j_i, k_i} ge 0$ and by the Prime Factorization Theorem we have $j_1 + j_2 + j_3 = 2$ and $k_1 + k_2 + k_3 = 3$.



          How many ways can you solve those equations, up to equivalent permutations of the factors? $12$



          Start with the first (smallest) factor $(2^0 3^0)$ and increment others for the first "row"; then $(2^1 3^0)$ and increment others for the second "row"; then $(2^0 3^1)$ and so on... for the third "row." There is no fourth "row"...



          Here are the sets, each in ascending order, themselves ordered:




          • {1, 1, 108}, {1, 2, 54}, {1, 3, 36}, {1, 4, 27}, {1, 6, 18}, {1, 9, 12}

          • {2, 2, 27}, {2, 3, 18}, {2, 6, 9}

          • {3, 3, 12}, {3, 4, 9}, {3, 6, 6}






          share|cite|improve this answer











          $endgroup$



          $$(2^{j_1} 3^{k_1})(2^{j_2}3^{k_2})(2^{j_3}3^{k_3}) = 108 = 2^2 cdot 3^3$$



          where ${ j_i, k_i} ge 0$ and by the Prime Factorization Theorem we have $j_1 + j_2 + j_3 = 2$ and $k_1 + k_2 + k_3 = 3$.



          How many ways can you solve those equations, up to equivalent permutations of the factors? $12$



          Start with the first (smallest) factor $(2^0 3^0)$ and increment others for the first "row"; then $(2^1 3^0)$ and increment others for the second "row"; then $(2^0 3^1)$ and so on... for the third "row." There is no fourth "row"...



          Here are the sets, each in ascending order, themselves ordered:




          • {1, 1, 108}, {1, 2, 54}, {1, 3, 36}, {1, 4, 27}, {1, 6, 18}, {1, 9, 12}

          • {2, 2, 27}, {2, 3, 18}, {2, 6, 9}

          • {3, 3, 12}, {3, 4, 9}, {3, 6, 6}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 15 at 7:44

























          answered Jan 15 at 7:19









          David G. StorkDavid G. Stork

          11k41432




          11k41432























              3












              $begingroup$

              $108=2^2 cdot 3^3$, so $x,y,z$ must be of the form $2^i3^j$ such that the sum of the $i$'s (where $0le ile 2$) is $2$ and the sum of the $j$'s (also integers starting from $0$) is $3$. How many solutions do $i_x+i_y+i_z = 2$ and $j_x+j_y+j_z = 3$ have in integers $ge 0$? All combinations of these solutions are the solutions you want.



              There are $binom{n+k-1}{k-1}$ non-negative integer solutions to $x_1+ ldots +x_k =n$ (see wikipedia, e.g.) and so for $3$ variables and $n=2$ we get $binom{2+3-1}{2} = binom{4}{2}=6$ and for $n=3$ we get $binom{5}{2} = 10$. So combined we get $6 times 10=60$ solutions for these independent equations.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                $108=2^2 cdot 3^3$, so $x,y,z$ must be of the form $2^i3^j$ such that the sum of the $i$'s (where $0le ile 2$) is $2$ and the sum of the $j$'s (also integers starting from $0$) is $3$. How many solutions do $i_x+i_y+i_z = 2$ and $j_x+j_y+j_z = 3$ have in integers $ge 0$? All combinations of these solutions are the solutions you want.



                There are $binom{n+k-1}{k-1}$ non-negative integer solutions to $x_1+ ldots +x_k =n$ (see wikipedia, e.g.) and so for $3$ variables and $n=2$ we get $binom{2+3-1}{2} = binom{4}{2}=6$ and for $n=3$ we get $binom{5}{2} = 10$. So combined we get $6 times 10=60$ solutions for these independent equations.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  $108=2^2 cdot 3^3$, so $x,y,z$ must be of the form $2^i3^j$ such that the sum of the $i$'s (where $0le ile 2$) is $2$ and the sum of the $j$'s (also integers starting from $0$) is $3$. How many solutions do $i_x+i_y+i_z = 2$ and $j_x+j_y+j_z = 3$ have in integers $ge 0$? All combinations of these solutions are the solutions you want.



                  There are $binom{n+k-1}{k-1}$ non-negative integer solutions to $x_1+ ldots +x_k =n$ (see wikipedia, e.g.) and so for $3$ variables and $n=2$ we get $binom{2+3-1}{2} = binom{4}{2}=6$ and for $n=3$ we get $binom{5}{2} = 10$. So combined we get $6 times 10=60$ solutions for these independent equations.






                  share|cite|improve this answer











                  $endgroup$



                  $108=2^2 cdot 3^3$, so $x,y,z$ must be of the form $2^i3^j$ such that the sum of the $i$'s (where $0le ile 2$) is $2$ and the sum of the $j$'s (also integers starting from $0$) is $3$. How many solutions do $i_x+i_y+i_z = 2$ and $j_x+j_y+j_z = 3$ have in integers $ge 0$? All combinations of these solutions are the solutions you want.



                  There are $binom{n+k-1}{k-1}$ non-negative integer solutions to $x_1+ ldots +x_k =n$ (see wikipedia, e.g.) and so for $3$ variables and $n=2$ we get $binom{2+3-1}{2} = binom{4}{2}=6$ and for $n=3$ we get $binom{5}{2} = 10$. So combined we get $6 times 10=60$ solutions for these independent equations.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 15 at 18:17

























                  answered Jan 15 at 7:12









                  Henno BrandsmaHenno Brandsma

                  110k347116




                  110k347116






























                      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%2f3074149%2ffind-the-cardinality-of-set-x-y-zx-y-z-in-mathbbz-wedge-xyz-108%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

                      Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                      Image
                      28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                      Read more

                      Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

                      Image
                      up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
                      Read more

                      A Topological Invariant for $pi_3(U(n))$

                      Image
                      3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
                      Read more