Find the smallest integer n such that the expression $40! cdot 5^n$ has the maximum number of trailing zeros.












0












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Find the smallest integer n such that the expression $$40! cdot 5^n$$ has the maximum number of
trailing zeros.






number-theory elementary-number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $text {Ans}=lfloor frac {40}{2} +frac {40}{4} +frac {40}{8} +frac {40}{16} +frac {40}{32} rfloor-lfloor frac {40}{5} +frac {40}{25} rfloor$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 16:17






  • 1




    $begingroup$
    @MohammadZuhairKhan Shouldn't the floor function arguments be kept separate before addition? It doesn't make a difference in this case, but it might in general.
    $endgroup$
    – Deepak
    Dec 14 '18 at 16:52


















0












$begingroup$


Find the smallest integer n such that the expression $$40! cdot 5^n$$ has the maximum number of
trailing zeros.






number-theory elementary-number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $text {Ans}=lfloor frac {40}{2} +frac {40}{4} +frac {40}{8} +frac {40}{16} +frac {40}{32} rfloor-lfloor frac {40}{5} +frac {40}{25} rfloor$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 16:17






  • 1




    $begingroup$
    @MohammadZuhairKhan Shouldn't the floor function arguments be kept separate before addition? It doesn't make a difference in this case, but it might in general.
    $endgroup$
    – Deepak
    Dec 14 '18 at 16:52
















0












0








0


2



$begingroup$


Find the smallest integer n such that the expression $$40! cdot 5^n$$ has the maximum number of
trailing zeros.






number-theory elementary-number-theory







share|cite|improve this question











$endgroup$




Find the smallest integer n such that the expression $$40! cdot 5^n$$ has the maximum number of
trailing zeros.






number-theory elementary-number-theory




number-theory elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 14 '18 at 16:09









Peter

46.9k1039125




46.9k1039125










asked Dec 14 '18 at 16:07









Isaiah LeobreraIsaiah Leobrera

301




301








  • 1




    $begingroup$
    $text {Ans}=lfloor frac {40}{2} +frac {40}{4} +frac {40}{8} +frac {40}{16} +frac {40}{32} rfloor-lfloor frac {40}{5} +frac {40}{25} rfloor$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 16:17






  • 1




    $begingroup$
    @MohammadZuhairKhan Shouldn't the floor function arguments be kept separate before addition? It doesn't make a difference in this case, but it might in general.
    $endgroup$
    – Deepak
    Dec 14 '18 at 16:52
















  • 1




    $begingroup$
    $text {Ans}=lfloor frac {40}{2} +frac {40}{4} +frac {40}{8} +frac {40}{16} +frac {40}{32} rfloor-lfloor frac {40}{5} +frac {40}{25} rfloor$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 16:17






  • 1




    $begingroup$
    @MohammadZuhairKhan Shouldn't the floor function arguments be kept separate before addition? It doesn't make a difference in this case, but it might in general.
    $endgroup$
    – Deepak
    Dec 14 '18 at 16:52










1




1




$begingroup$
$text {Ans}=lfloor frac {40}{2} +frac {40}{4} +frac {40}{8} +frac {40}{16} +frac {40}{32} rfloor-lfloor frac {40}{5} +frac {40}{25} rfloor$
$endgroup$
– Mohammad Zuhair Khan
Dec 14 '18 at 16:17




$begingroup$
$text {Ans}=lfloor frac {40}{2} +frac {40}{4} +frac {40}{8} +frac {40}{16} +frac {40}{32} rfloor-lfloor frac {40}{5} +frac {40}{25} rfloor$
$endgroup$
– Mohammad Zuhair Khan
Dec 14 '18 at 16:17




1




1




$begingroup$
@MohammadZuhairKhan Shouldn't the floor function arguments be kept separate before addition? It doesn't make a difference in this case, but it might in general.
$endgroup$
– Deepak
Dec 14 '18 at 16:52






$begingroup$
@MohammadZuhairKhan Shouldn't the floor function arguments be kept separate before addition? It doesn't make a difference in this case, but it might in general.
$endgroup$
– Deepak
Dec 14 '18 at 16:52












2 Answers
2






active

oldest

votes


















2












$begingroup$

Hint: Legendre's theorem
can be used to find the largest powers of $2$ and of $5$ that divide $40!$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Basically, you need to count the number of factors $2$ and $5$ in the factorial $40!$:
    $$begin{align}40!&=color{red}1cdot 2cdot color{red}3cdot 4cdot color{red}5cdots 38cdot color{red}{39}cdot 40=\
    &=2^{20}cdot (color{blue}1cdot 2cdot color{blue}3cdots color{blue}{19}cdot 20)cdot (color{red}1cdot color{red}3cdot color{red}5cdots color{red}{37}cdot color{red}{39})=\
    &=2^{20}cdot 2^{10}cdot (1cdot 2cdot 3cdots 9cdot 10)cdot (color{blue}1cdot color{blue}3cdots color{blue}{19})cdot (1cdot 3cdot 5cdots 37cdot 39)=\
    &=2^{20}cdot 2^{10}cdot 2^5cdot (1cdot 2cdot 3cdot 4cdot 5)cdot (1cdot 3cdots 9)cdot (1cdot 3cdots 19)cdot (1cdot 3cdot 5cdots 37cdot 39)=\
    &=2^{20}cdot 2^{10}cdot 2^5cdot 2^3cdot (1cdot 3cdot 5)cdot (1cdots 9)cdot (1cdots 19)cdot (1cdots 39)=\
    &=2^{38}cdot (5A)cdot (5B)cdot (5^2C)cdot (5^5D)=\
    &=2^{38}cdot 5^9E. end{align}$$
    Hence, the minimum $n$ in $40!cdot 5^n$ is: $38=9+nRightarrow n=29$.






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      2 Answers
      2






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      2 Answers
      2






      active

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      active

      oldest

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      active

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      2












      $begingroup$

      Hint: Legendre's theorem
      can be used to find the largest powers of $2$ and of $5$ that divide $40!$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Hint: Legendre's theorem
        can be used to find the largest powers of $2$ and of $5$ that divide $40!$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Hint: Legendre's theorem
          can be used to find the largest powers of $2$ and of $5$ that divide $40!$.






          share|cite|improve this answer









          $endgroup$



          Hint: Legendre's theorem
          can be used to find the largest powers of $2$ and of $5$ that divide $40!$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 16:14









          Robert IsraelRobert Israel

          319k23209460




          319k23209460























              1












              $begingroup$

              Basically, you need to count the number of factors $2$ and $5$ in the factorial $40!$:
              $$begin{align}40!&=color{red}1cdot 2cdot color{red}3cdot 4cdot color{red}5cdots 38cdot color{red}{39}cdot 40=\
              &=2^{20}cdot (color{blue}1cdot 2cdot color{blue}3cdots color{blue}{19}cdot 20)cdot (color{red}1cdot color{red}3cdot color{red}5cdots color{red}{37}cdot color{red}{39})=\
              &=2^{20}cdot 2^{10}cdot (1cdot 2cdot 3cdots 9cdot 10)cdot (color{blue}1cdot color{blue}3cdots color{blue}{19})cdot (1cdot 3cdot 5cdots 37cdot 39)=\
              &=2^{20}cdot 2^{10}cdot 2^5cdot (1cdot 2cdot 3cdot 4cdot 5)cdot (1cdot 3cdots 9)cdot (1cdot 3cdots 19)cdot (1cdot 3cdot 5cdots 37cdot 39)=\
              &=2^{20}cdot 2^{10}cdot 2^5cdot 2^3cdot (1cdot 3cdot 5)cdot (1cdots 9)cdot (1cdots 19)cdot (1cdots 39)=\
              &=2^{38}cdot (5A)cdot (5B)cdot (5^2C)cdot (5^5D)=\
              &=2^{38}cdot 5^9E. end{align}$$
              Hence, the minimum $n$ in $40!cdot 5^n$ is: $38=9+nRightarrow n=29$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Basically, you need to count the number of factors $2$ and $5$ in the factorial $40!$:
                $$begin{align}40!&=color{red}1cdot 2cdot color{red}3cdot 4cdot color{red}5cdots 38cdot color{red}{39}cdot 40=\
                &=2^{20}cdot (color{blue}1cdot 2cdot color{blue}3cdots color{blue}{19}cdot 20)cdot (color{red}1cdot color{red}3cdot color{red}5cdots color{red}{37}cdot color{red}{39})=\
                &=2^{20}cdot 2^{10}cdot (1cdot 2cdot 3cdots 9cdot 10)cdot (color{blue}1cdot color{blue}3cdots color{blue}{19})cdot (1cdot 3cdot 5cdots 37cdot 39)=\
                &=2^{20}cdot 2^{10}cdot 2^5cdot (1cdot 2cdot 3cdot 4cdot 5)cdot (1cdot 3cdots 9)cdot (1cdot 3cdots 19)cdot (1cdot 3cdot 5cdots 37cdot 39)=\
                &=2^{20}cdot 2^{10}cdot 2^5cdot 2^3cdot (1cdot 3cdot 5)cdot (1cdots 9)cdot (1cdots 19)cdot (1cdots 39)=\
                &=2^{38}cdot (5A)cdot (5B)cdot (5^2C)cdot (5^5D)=\
                &=2^{38}cdot 5^9E. end{align}$$
                Hence, the minimum $n$ in $40!cdot 5^n$ is: $38=9+nRightarrow n=29$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Basically, you need to count the number of factors $2$ and $5$ in the factorial $40!$:
                  $$begin{align}40!&=color{red}1cdot 2cdot color{red}3cdot 4cdot color{red}5cdots 38cdot color{red}{39}cdot 40=\
                  &=2^{20}cdot (color{blue}1cdot 2cdot color{blue}3cdots color{blue}{19}cdot 20)cdot (color{red}1cdot color{red}3cdot color{red}5cdots color{red}{37}cdot color{red}{39})=\
                  &=2^{20}cdot 2^{10}cdot (1cdot 2cdot 3cdots 9cdot 10)cdot (color{blue}1cdot color{blue}3cdots color{blue}{19})cdot (1cdot 3cdot 5cdots 37cdot 39)=\
                  &=2^{20}cdot 2^{10}cdot 2^5cdot (1cdot 2cdot 3cdot 4cdot 5)cdot (1cdot 3cdots 9)cdot (1cdot 3cdots 19)cdot (1cdot 3cdot 5cdots 37cdot 39)=\
                  &=2^{20}cdot 2^{10}cdot 2^5cdot 2^3cdot (1cdot 3cdot 5)cdot (1cdots 9)cdot (1cdots 19)cdot (1cdots 39)=\
                  &=2^{38}cdot (5A)cdot (5B)cdot (5^2C)cdot (5^5D)=\
                  &=2^{38}cdot 5^9E. end{align}$$
                  Hence, the minimum $n$ in $40!cdot 5^n$ is: $38=9+nRightarrow n=29$.






                  share|cite|improve this answer









                  $endgroup$



                  Basically, you need to count the number of factors $2$ and $5$ in the factorial $40!$:
                  $$begin{align}40!&=color{red}1cdot 2cdot color{red}3cdot 4cdot color{red}5cdots 38cdot color{red}{39}cdot 40=\
                  &=2^{20}cdot (color{blue}1cdot 2cdot color{blue}3cdots color{blue}{19}cdot 20)cdot (color{red}1cdot color{red}3cdot color{red}5cdots color{red}{37}cdot color{red}{39})=\
                  &=2^{20}cdot 2^{10}cdot (1cdot 2cdot 3cdots 9cdot 10)cdot (color{blue}1cdot color{blue}3cdots color{blue}{19})cdot (1cdot 3cdot 5cdots 37cdot 39)=\
                  &=2^{20}cdot 2^{10}cdot 2^5cdot (1cdot 2cdot 3cdot 4cdot 5)cdot (1cdot 3cdots 9)cdot (1cdot 3cdots 19)cdot (1cdot 3cdot 5cdots 37cdot 39)=\
                  &=2^{20}cdot 2^{10}cdot 2^5cdot 2^3cdot (1cdot 3cdot 5)cdot (1cdots 9)cdot (1cdots 19)cdot (1cdots 39)=\
                  &=2^{38}cdot (5A)cdot (5B)cdot (5^2C)cdot (5^5D)=\
                  &=2^{38}cdot 5^9E. end{align}$$
                  Hence, the minimum $n$ in $40!cdot 5^n$ is: $38=9+nRightarrow n=29$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 14 '18 at 17:01









                  farruhotafarruhota

                  19.6k2738




                  19.6k2738






























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