Find the smallest integer n such that the expression $40! cdot 5^n$ has the maximum number of trailing zeros.
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
add a comment |
$begingroup$
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
$endgroup$
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
edited Dec 14 '18 at 16:09
Peter
47.6k1039131
47.6k1039131
asked Dec 14 '18 at 16:07
Isaiah LeobreraIsaiah Leobrera
281
281
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
add a comment |
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
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Hint: Legendre's theorem
can be used to find the largest powers of $2$ and of $5$ that divide $40!$.
$endgroup$
add a comment |
$begingroup$
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: $38=9+nRightarrow n=29$.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Legendre's theorem
can be used to find the largest powers of $2$ and of $5$ that divide $40!$.
$endgroup$
add a comment |
$begingroup$
Hint: Legendre's theorem
can be used to find the largest powers of $2$ and of $5$ that divide $40!$.
$endgroup$
add a comment |
$begingroup$
Hint: Legendre's theorem
can be used to find the largest powers of $2$ and of $5$ that divide $40!$.
$endgroup$
Hint: Legendre's theorem
can be used to find the largest powers of $2$ and of $5$ that divide $40!$.
answered Dec 14 '18 at 16:14
Robert IsraelRobert Israel
323k23213467
323k23213467
add a comment |
add a comment |
$begingroup$
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: $38=9+nRightarrow n=29$.
$endgroup$
add a comment |
$begingroup$
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: $38=9+nRightarrow n=29$.
$endgroup$
add a comment |
$begingroup$
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: $38=9+nRightarrow n=29$.
$endgroup$
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: $38=9+nRightarrow n=29$.
answered Dec 14 '18 at 17:01
farruhotafarruhota
20.4k2739
20.4k2739
add a comment |
add a comment |
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$begingroup$
$text {Ans}=lfloor frac {40}{2} +frac {40}{4} +frac {40}{8} +frac {40}{16} +frac {40}{32} rfloor-lfloor frac {40}{5} +frac {40}{25} rfloor$
$endgroup$
– 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