factorial

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simu
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factorial

Unread post by simu » Wed Jan 16, 2013 7:04 pm

যদি n!= 40320 হয়, তাহলে n= ?

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sowmitra
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Re: factorial

Unread post by sowmitra » Wed Jan 16, 2013 9:06 pm

The prime factorization of $40320$ is :
\[\displaystyle40320=2^7\times3^2\times5\times7\]
Here, you can see that $7$ and $5$ occur only once in the factorization. So, $40320$ can be at least $7!$ and at most $9!$. But, in $9!$, $3$ occurs $3$ times. So, $n\neq9$. Now, if you check,
$\displaystyle7!=5040$
$\displaystyle8!=40320$
So, $\boxed{n=8}$.
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nafistiham
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Re: factorial

Unread post by nafistiham » Wed Jan 16, 2013 10:24 pm

sowmitra wrote:The prime factorization of $40320$ is :
\[\displaystyle40320=2^7\times3^2\times5\times7\]
Here, you can see that $7$ and $5$ occur only once in the factorization. So, $40320$ can be at least $7!$ and at most $9!$. But, in $9!$, $3$ occurs $3$ times. So, $n\neq9$. Now, if you check,
$\displaystyle7!=5040$
$\displaystyle8!=40320$
So, $\boxed{n=8}$.
That's quite smart. But, if at last, you are dividing (and ruling :D :D ) the number to its prime factors, why not divide just $7$ times ?
divide it by $1,2,3,4,5,6....$ serially.
So, you need less time, less calculation and, even less brain :lol:
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Phlembac Adib Hasan
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Re: factorial

Unread post by Phlembac Adib Hasan » Thu Jan 17, 2013 1:21 pm

sowmitra wrote:The prime factorization of $40320$ is :
\[\displaystyle40320=2^7\times3^2\times5\times7\]
Here, you can see that $7$ and $5$ occur only once in the factorization. So, $40320$ can be at least $7!$ and at most $9!$. But, in $9!$, $3$ occurs $3$ times. So, $n\neq9$. Now, if you check,
$\displaystyle7!=5040$
$\displaystyle8!=40320$
So, $\boxed{n=8}$.
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আমার সমাধান দিচ্ছি।
After factorization, we find $2^7||40320$. Now note that $2^4||7!$ and $2^8||10!$. So $n=8$ or $n=9$. The second one is impossible because $3^4||9!$ where $3^2||40320$. So $n=8$.
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