One term of (2π+5)^n = 288000π^8, what's n?












0














Without using calculator what's the value of n?



Using binomial expansion I get:



nCp * 2n-p * πn-p * 5p = 288000π8



Easily I know that n-p=8, by the π's power



Then the power of 2 is also 8, so I can divide both sides



nCp * 5p = 1125



Well I know that




  • n-p=8

  • n-8=p

  • n=p+8


And with "pascal triangle's" rule(nCp = nCn-p)



nC8 * 5p = 1125



nC8 * 5n-8 = 1125



Well there are definitively many ways to equate this, but how can I get the value of n without calculator or guessing(Trial and error)?



Or is there any other approach?



The answer is n=10 and p=2...



Sorry for no LaTeX(I don't know how to work with it






combinatorics binomial-coefficients binomial-theorem







share|cite|improve this question




















  • 1




    math.meta.stackexchange.com/questions/5020/…
    – saulspatz
    Nov 29 '18 at 22:12






  • 1




    The initial use of binomial theorem doesn't look quite right -- you'd want $pi^p$ instead of $pi^{(n-p) * p}$. So either $p = 8$ or $n-p = 8$ and you would have $2^p$ instead of $2^{n-p}$.
    – platty
    Nov 29 '18 at 22:19










  • It was my bad sorry, corrected it
    – Nuno Mateus
    Nov 29 '18 at 22:23
















0














Without using calculator what's the value of n?



Using binomial expansion I get:



nCp * 2n-p * πn-p * 5p = 288000π8



Easily I know that n-p=8, by the π's power



Then the power of 2 is also 8, so I can divide both sides



nCp * 5p = 1125



Well I know that




  • n-p=8

  • n-8=p

  • n=p+8


And with "pascal triangle's" rule(nCp = nCn-p)



nC8 * 5p = 1125



nC8 * 5n-8 = 1125



Well there are definitively many ways to equate this, but how can I get the value of n without calculator or guessing(Trial and error)?



Or is there any other approach?



The answer is n=10 and p=2...



Sorry for no LaTeX(I don't know how to work with it






combinatorics binomial-coefficients binomial-theorem







share|cite|improve this question




















  • 1




    math.meta.stackexchange.com/questions/5020/…
    – saulspatz
    Nov 29 '18 at 22:12






  • 1




    The initial use of binomial theorem doesn't look quite right -- you'd want $pi^p$ instead of $pi^{(n-p) * p}$. So either $p = 8$ or $n-p = 8$ and you would have $2^p$ instead of $2^{n-p}$.
    – platty
    Nov 29 '18 at 22:19










  • It was my bad sorry, corrected it
    – Nuno Mateus
    Nov 29 '18 at 22:23














0












0








0







Without using calculator what's the value of n?



Using binomial expansion I get:



nCp * 2n-p * πn-p * 5p = 288000π8



Easily I know that n-p=8, by the π's power



Then the power of 2 is also 8, so I can divide both sides



nCp * 5p = 1125



Well I know that




  • n-p=8

  • n-8=p

  • n=p+8


And with "pascal triangle's" rule(nCp = nCn-p)



nC8 * 5p = 1125



nC8 * 5n-8 = 1125



Well there are definitively many ways to equate this, but how can I get the value of n without calculator or guessing(Trial and error)?



Or is there any other approach?



The answer is n=10 and p=2...



Sorry for no LaTeX(I don't know how to work with it






combinatorics binomial-coefficients binomial-theorem







share|cite|improve this question















Without using calculator what's the value of n?



Using binomial expansion I get:



nCp * 2n-p * πn-p * 5p = 288000π8



Easily I know that n-p=8, by the π's power



Then the power of 2 is also 8, so I can divide both sides



nCp * 5p = 1125



Well I know that




  • n-p=8

  • n-8=p

  • n=p+8


And with "pascal triangle's" rule(nCp = nCn-p)



nC8 * 5p = 1125



nC8 * 5n-8 = 1125



Well there are definitively many ways to equate this, but how can I get the value of n without calculator or guessing(Trial and error)?



Or is there any other approach?



The answer is n=10 and p=2...



Sorry for no LaTeX(I don't know how to work with it






combinatorics binomial-coefficients binomial-theorem




combinatorics binomial-coefficients binomial-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 22:23







Nuno Mateus

















asked Nov 29 '18 at 22:07









Nuno MateusNuno Mateus

494




494








  • 1




    math.meta.stackexchange.com/questions/5020/…
    – saulspatz
    Nov 29 '18 at 22:12






  • 1




    The initial use of binomial theorem doesn't look quite right -- you'd want $pi^p$ instead of $pi^{(n-p) * p}$. So either $p = 8$ or $n-p = 8$ and you would have $2^p$ instead of $2^{n-p}$.
    – platty
    Nov 29 '18 at 22:19










  • It was my bad sorry, corrected it
    – Nuno Mateus
    Nov 29 '18 at 22:23














  • 1




    math.meta.stackexchange.com/questions/5020/…
    – saulspatz
    Nov 29 '18 at 22:12






  • 1




    The initial use of binomial theorem doesn't look quite right -- you'd want $pi^p$ instead of $pi^{(n-p) * p}$. So either $p = 8$ or $n-p = 8$ and you would have $2^p$ instead of $2^{n-p}$.
    – platty
    Nov 29 '18 at 22:19










  • It was my bad sorry, corrected it
    – Nuno Mateus
    Nov 29 '18 at 22:23








1




1




math.meta.stackexchange.com/questions/5020/…
– saulspatz
Nov 29 '18 at 22:12




math.meta.stackexchange.com/questions/5020/…
– saulspatz
Nov 29 '18 at 22:12




1




1




The initial use of binomial theorem doesn't look quite right -- you'd want $pi^p$ instead of $pi^{(n-p) * p}$. So either $p = 8$ or $n-p = 8$ and you would have $2^p$ instead of $2^{n-p}$.
– platty
Nov 29 '18 at 22:19




The initial use of binomial theorem doesn't look quite right -- you'd want $pi^p$ instead of $pi^{(n-p) * p}$. So either $p = 8$ or $n-p = 8$ and you would have $2^p$ instead of $2^{n-p}$.
– platty
Nov 29 '18 at 22:19












It was my bad sorry, corrected it
– Nuno Mateus
Nov 29 '18 at 22:23




It was my bad sorry, corrected it
– Nuno Mateus
Nov 29 '18 at 22:23










2 Answers
2






active

oldest

votes


















3














The term of degree $8$ in the expansion of $(x+a)^n$ is
$$
binom{n}{8}a^{n-8}x^8
$$
so you want to solve
$$
binom{n}{8}5^{n-8}2^8=288000=2^83^25^3
$$
hence
$$
binom{n}{8}5^{n-11}=9
$$
We can exclude $n>11$, because the right hand side is not divisible by $5$.



Therefore you just have to check $n=8,9,10,11$:
begin{align}
binom{8}{8}&=1 \
binom{9}{8}&=binom{9}{1}=9 \
binom{10}{8}&=binom{10}{2}=cdots\
binom{11}{8}&=binom{11}{3}=cdots
end{align}






share|cite|improve this answer





























    1















    ${nchoose 8} * 5^{n-8} = 1125$




    But $1125 = 9*5^3$ so $5^{n-8}|5^3$ and $8le n le 11$



    And $5^{n-11}{nchoose 8} = 9$.



    $9not mid 5$ so either $n-11=0$ or $5^{11-n}|{nchoose 8}$.



    Now as $8le nle 11$ and ${nchoose 8}=frac {9*..n}{(n-8)!}$ then $11 - n$ is at most $1$.



    So $n = 10$ or $11$. Now as no prime but $3,5$ divide $frac {9*..*n}{(n-8)!}$ we can not have $n = 11$ so



    so $n =10$.






    share|cite|improve this answer























      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%2f3019301%2fone-term-of-2%25cf%25805n-288000%25cf%25808-whats-n%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














      The term of degree $8$ in the expansion of $(x+a)^n$ is
      $$
      binom{n}{8}a^{n-8}x^8
      $$
      so you want to solve
      $$
      binom{n}{8}5^{n-8}2^8=288000=2^83^25^3
      $$
      hence
      $$
      binom{n}{8}5^{n-11}=9
      $$
      We can exclude $n>11$, because the right hand side is not divisible by $5$.



      Therefore you just have to check $n=8,9,10,11$:
      begin{align}
      binom{8}{8}&=1 \
      binom{9}{8}&=binom{9}{1}=9 \
      binom{10}{8}&=binom{10}{2}=cdots\
      binom{11}{8}&=binom{11}{3}=cdots
      end{align}






      share|cite|improve this answer


























        3














        The term of degree $8$ in the expansion of $(x+a)^n$ is
        $$
        binom{n}{8}a^{n-8}x^8
        $$
        so you want to solve
        $$
        binom{n}{8}5^{n-8}2^8=288000=2^83^25^3
        $$
        hence
        $$
        binom{n}{8}5^{n-11}=9
        $$
        We can exclude $n>11$, because the right hand side is not divisible by $5$.



        Therefore you just have to check $n=8,9,10,11$:
        begin{align}
        binom{8}{8}&=1 \
        binom{9}{8}&=binom{9}{1}=9 \
        binom{10}{8}&=binom{10}{2}=cdots\
        binom{11}{8}&=binom{11}{3}=cdots
        end{align}






        share|cite|improve this answer
























          3












          3








          3






          The term of degree $8$ in the expansion of $(x+a)^n$ is
          $$
          binom{n}{8}a^{n-8}x^8
          $$
          so you want to solve
          $$
          binom{n}{8}5^{n-8}2^8=288000=2^83^25^3
          $$
          hence
          $$
          binom{n}{8}5^{n-11}=9
          $$
          We can exclude $n>11$, because the right hand side is not divisible by $5$.



          Therefore you just have to check $n=8,9,10,11$:
          begin{align}
          binom{8}{8}&=1 \
          binom{9}{8}&=binom{9}{1}=9 \
          binom{10}{8}&=binom{10}{2}=cdots\
          binom{11}{8}&=binom{11}{3}=cdots
          end{align}






          share|cite|improve this answer












          The term of degree $8$ in the expansion of $(x+a)^n$ is
          $$
          binom{n}{8}a^{n-8}x^8
          $$
          so you want to solve
          $$
          binom{n}{8}5^{n-8}2^8=288000=2^83^25^3
          $$
          hence
          $$
          binom{n}{8}5^{n-11}=9
          $$
          We can exclude $n>11$, because the right hand side is not divisible by $5$.



          Therefore you just have to check $n=8,9,10,11$:
          begin{align}
          binom{8}{8}&=1 \
          binom{9}{8}&=binom{9}{1}=9 \
          binom{10}{8}&=binom{10}{2}=cdots\
          binom{11}{8}&=binom{11}{3}=cdots
          end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 22:39









          egregegreg

          179k1485202




          179k1485202























              1















              ${nchoose 8} * 5^{n-8} = 1125$




              But $1125 = 9*5^3$ so $5^{n-8}|5^3$ and $8le n le 11$



              And $5^{n-11}{nchoose 8} = 9$.



              $9not mid 5$ so either $n-11=0$ or $5^{11-n}|{nchoose 8}$.



              Now as $8le nle 11$ and ${nchoose 8}=frac {9*..n}{(n-8)!}$ then $11 - n$ is at most $1$.



              So $n = 10$ or $11$. Now as no prime but $3,5$ divide $frac {9*..*n}{(n-8)!}$ we can not have $n = 11$ so



              so $n =10$.






              share|cite|improve this answer




























                1















                ${nchoose 8} * 5^{n-8} = 1125$




                But $1125 = 9*5^3$ so $5^{n-8}|5^3$ and $8le n le 11$



                And $5^{n-11}{nchoose 8} = 9$.



                $9not mid 5$ so either $n-11=0$ or $5^{11-n}|{nchoose 8}$.



                Now as $8le nle 11$ and ${nchoose 8}=frac {9*..n}{(n-8)!}$ then $11 - n$ is at most $1$.



                So $n = 10$ or $11$. Now as no prime but $3,5$ divide $frac {9*..*n}{(n-8)!}$ we can not have $n = 11$ so



                so $n =10$.






                share|cite|improve this answer


























                  1












                  1








                  1







                  ${nchoose 8} * 5^{n-8} = 1125$




                  But $1125 = 9*5^3$ so $5^{n-8}|5^3$ and $8le n le 11$



                  And $5^{n-11}{nchoose 8} = 9$.



                  $9not mid 5$ so either $n-11=0$ or $5^{11-n}|{nchoose 8}$.



                  Now as $8le nle 11$ and ${nchoose 8}=frac {9*..n}{(n-8)!}$ then $11 - n$ is at most $1$.



                  So $n = 10$ or $11$. Now as no prime but $3,5$ divide $frac {9*..*n}{(n-8)!}$ we can not have $n = 11$ so



                  so $n =10$.






                  share|cite|improve this answer















                  ${nchoose 8} * 5^{n-8} = 1125$




                  But $1125 = 9*5^3$ so $5^{n-8}|5^3$ and $8le n le 11$



                  And $5^{n-11}{nchoose 8} = 9$.



                  $9not mid 5$ so either $n-11=0$ or $5^{11-n}|{nchoose 8}$.



                  Now as $8le nle 11$ and ${nchoose 8}=frac {9*..n}{(n-8)!}$ then $11 - n$ is at most $1$.



                  So $n = 10$ or $11$. Now as no prime but $3,5$ divide $frac {9*..*n}{(n-8)!}$ we can not have $n = 11$ so



                  so $n =10$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 30 '18 at 0:14

























                  answered Nov 30 '18 at 0:04









                  fleabloodfleablood

                  68.7k22685




                  68.7k22685






























                      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%2f3019301%2fone-term-of-2%25cf%25805n-288000%25cf%25808-whats-n%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

                      Quarter-circle Tiles

                      Image
                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
                      Read more

                      build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

                      Image
                      0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
                      Read more

                      Mont Emei

                      Image
                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
                      Read more