Prove that $binom{2n}{n}$ is not divisible by $p$
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Let $n$ be an integer greater than $5$. I would like to prove that if $p$ is a prime such that $ frac{2}{3}n < p leq n$ then $binom{2n}{n}$ is not a multiple of $p$.
I think I could assume that $binom{2n}{n}$ is a multiple of $p$ but I do not really know what to make of it.
elementary-number-theory binomial-coefficients divisibility
edited Jan 11 at 21:11
Lorenzo B.
1,8402520
asked Sep 14 '15 at 17:02
OdileOdile
62648
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add a comment |
$begingroup$
Let $n$ be an integer greater than $5$. I would like to prove that if $p$ is a prime such that $ frac{2}{3}n < p leq n$ then $binom{2n}{n}$ is not a multiple of $p$.
I think I could assume that $binom{2n}{n}$ is a multiple of $p$ but I do not really know what to make of it.
elementary-number-theory binomial-coefficients divisibility
edited Jan 11 at 21:11
Lorenzo B.
1,8402520
asked Sep 14 '15 at 17:02
OdileOdile
62648
$endgroup$
add a comment |
$begingroup$
Let $n$ be an integer greater than $5$. I would like to prove that if $p$ is a prime such that $ frac{2}{3}n < p leq n$ then $binom{2n}{n}$ is not a multiple of $p$.
I think I could assume that $binom{2n}{n}$ is a multiple of $p$ but I do not really know what to make of it.
elementary-number-theory binomial-coefficients divisibility
edited Jan 11 at 21:11
Lorenzo B.
1,8402520
asked Sep 14 '15 at 17:02
OdileOdile
62648
$endgroup$
Let $n$ be an integer greater than $5$. I would like to prove that if $p$ is a prime such that $ frac{2}{3}n < p leq n$ then $binom{2n}{n}$ is not a multiple of $p$.
I think I could assume that $binom{2n}{n}$ is a multiple of $p$ but I do not really know what to make of it.
elementary-number-theory binomial-coefficients divisibility
elementary-number-theory binomial-coefficients divisibility
edited Jan 11 at 21:11
Lorenzo B.
1,8402520
asked Sep 14 '15 at 17:02
OdileOdile
62648
edited Jan 11 at 21:11
Lorenzo B.
1,8402520
asked Sep 14 '15 at 17:02
OdileOdile
62648
edited Jan 11 at 21:11
Lorenzo B.
1,8402520
edited Jan 11 at 21:11
Lorenzo B.
1,8402520
edited Jan 11 at 21:11
Lorenzo B.
1,8402520
1,8402520
asked Sep 14 '15 at 17:02
OdileOdile
62648
asked Sep 14 '15 at 17:02
OdileOdile
62648
asked Sep 14 '15 at 17:02
OdileOdile
62648
62648
add a comment |
add a comment |
1 Answer
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Recall that $${2nchoose n}=frac{(2n)!}{n!n!} $$
and that the exponent $k=v_p(m!)$ of $p$ such that $p^kmid m!$ and $p^{k+1}mid m!$ is given by
$$ v_p(m!)=leftlfloor frac mprightrfloor+leftlfloor frac m{p^2}rightrfloor+leftlfloor frac m{p^3}rightrfloor+ldots$$
From the given conditions, $v_p(n!)=1+0+ldots =1$ and $v_p((2n)!)=2+lfloor frac {2}{p}rfloor+ldots$, which can only be $>2$ if $p=2$, but $p>frac23nge frac{10}3>2$. Hence $v_p$ of the binomial is $2-1-1=0$.
answered Sep 14 '15 at 17:15
Hagen von EitzenHagen von Eitzen
278k22269497
$endgroup$
$begingroup$
Great answer ! Thank you !!
$endgroup$
– Odile
Sep 14 '15 at 17:27
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Recall that $${2nchoose n}=frac{(2n)!}{n!n!} $$
and that the exponent $k=v_p(m!)$ of $p$ such that $p^kmid m!$ and $p^{k+1}mid m!$ is given by
$$ v_p(m!)=leftlfloor frac mprightrfloor+leftlfloor frac m{p^2}rightrfloor+leftlfloor frac m{p^3}rightrfloor+ldots$$
From the given conditions, $v_p(n!)=1+0+ldots =1$ and $v_p((2n)!)=2+lfloor frac {2}{p}rfloor+ldots$, which can only be $>2$ if $p=2$, but $p>frac23nge frac{10}3>2$. Hence $v_p$ of the binomial is $2-1-1=0$.
answered Sep 14 '15 at 17:15
Hagen von EitzenHagen von Eitzen
278k22269497
$endgroup$
$begingroup$
Great answer ! Thank you !!
$endgroup$
– Odile
Sep 14 '15 at 17:27
add a comment |
$begingroup$
Recall that $${2nchoose n}=frac{(2n)!}{n!n!} $$
and that the exponent $k=v_p(m!)$ of $p$ such that $p^kmid m!$ and $p^{k+1}mid m!$ is given by
$$ v_p(m!)=leftlfloor frac mprightrfloor+leftlfloor frac m{p^2}rightrfloor+leftlfloor frac m{p^3}rightrfloor+ldots$$
From the given conditions, $v_p(n!)=1+0+ldots =1$ and $v_p((2n)!)=2+lfloor frac {2}{p}rfloor+ldots$, which can only be $>2$ if $p=2$, but $p>frac23nge frac{10}3>2$. Hence $v_p$ of the binomial is $2-1-1=0$.
answered Sep 14 '15 at 17:15
Hagen von EitzenHagen von Eitzen
278k22269497
$endgroup$
$begingroup$
Great answer ! Thank you !!
$endgroup$
– Odile
Sep 14 '15 at 17:27
add a comment |
$begingroup$
Recall that $${2nchoose n}=frac{(2n)!}{n!n!} $$
and that the exponent $k=v_p(m!)$ of $p$ such that $p^kmid m!$ and $p^{k+1}mid m!$ is given by
$$ v_p(m!)=leftlfloor frac mprightrfloor+leftlfloor frac m{p^2}rightrfloor+leftlfloor frac m{p^3}rightrfloor+ldots$$
From the given conditions, $v_p(n!)=1+0+ldots =1$ and $v_p((2n)!)=2+lfloor frac {2}{p}rfloor+ldots$, which can only be $>2$ if $p=2$, but $p>frac23nge frac{10}3>2$. Hence $v_p$ of the binomial is $2-1-1=0$.
answered Sep 14 '15 at 17:15
Hagen von EitzenHagen von Eitzen
278k22269497
$endgroup$
Recall that $${2nchoose n}=frac{(2n)!}{n!n!} $$
and that the exponent $k=v_p(m!)$ of $p$ such that $p^kmid m!$ and $p^{k+1}mid m!$ is given by
$$ v_p(m!)=leftlfloor frac mprightrfloor+leftlfloor frac m{p^2}rightrfloor+leftlfloor frac m{p^3}rightrfloor+ldots$$
From the given conditions, $v_p(n!)=1+0+ldots =1$ and $v_p((2n)!)=2+lfloor frac {2}{p}rfloor+ldots$, which can only be $>2$ if $p=2$, but $p>frac23nge frac{10}3>2$. Hence $v_p$ of the binomial is $2-1-1=0$.
answered Sep 14 '15 at 17:15
Hagen von EitzenHagen von Eitzen
278k22269497
answered Sep 14 '15 at 17:15
Hagen von EitzenHagen von Eitzen
278k22269497
answered Sep 14 '15 at 17:15
Hagen von EitzenHagen von Eitzen
278k22269497
answered Sep 14 '15 at 17:15
Hagen von EitzenHagen von Eitzen
278k22269497
278k22269497
$begingroup$
Great answer ! Thank you !!
$endgroup$
– Odile
Sep 14 '15 at 17:27
add a comment |
$begingroup$
Great answer ! Thank you !!
$endgroup$
– Odile
Sep 14 '15 at 17:27
$begingroup$
Great answer ! Thank you !!
$endgroup$
– Odile
Sep 14 '15 at 17:27
$begingroup$
Great answer ! Thank you !!
$endgroup$
– Odile
Sep 14 '15 at 17:27
add a comment |
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