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












0












$begingroup$


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











$endgroup$








  • 1




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






  • 1




    $begingroup$
    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 $2^{n-p}$.
    $endgroup$
    – platty
    Nov 29 '18 at 22:19










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
















0












$begingroup$


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











$endgroup$








  • 1




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






  • 1




    $begingroup$
    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 $2^{n-p}$.
    $endgroup$
    – platty
    Nov 29 '18 at 22:19










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














0












0








0





$begingroup$


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











$endgroup$




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




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






  • 1




    $begingroup$
    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 $2^{n-p}$.
    $endgroup$
    – platty
    Nov 29 '18 at 22:19










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














  • 1




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






  • 1




    $begingroup$
    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 $2^{n-p}$.
    $endgroup$
    – platty
    Nov 29 '18 at 22:19










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








1




1




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




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




1




1




$begingroup$
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 $2^{n-p}$.
$endgroup$
– platty
Nov 29 '18 at 22:19




$begingroup$
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 $2^{n-p}$.
$endgroup$
– platty
Nov 29 '18 at 22:19












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




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










2 Answers
2






active

oldest

votes


















3












$begingroup$

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









$endgroup$





















    1












    $begingroup$


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




    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











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      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









      $endgroup$


















        3












        $begingroup$

        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









        $endgroup$
















          3












          3








          3





          $begingroup$

          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









          $endgroup$



          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

          180k1485202




          180k1485202























              1












              $begingroup$


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




              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











              $endgroup$


















                1












                $begingroup$


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




                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











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$


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




                  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











                  $endgroup$




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




                  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

                  68.9k22685




                  68.9k22685






























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