## factorial

### Re: factorial

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, $\boxed{n=8}$.

\[\displaystyle40320=2^7\times3^2\times5\times7\]

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, $\boxed{n=8}$.

- nafistiham
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### Re: factorial

That's quite smart. But, if at last, you are dividing (and ruling ) the number to its prime factors, why not divide just $7$ times ?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, $\boxed{n=8}$.

divide it by $1,2,3,4,5,6....$ serially.

So, you need less time, less calculation and, even less brain

\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]

Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

- Phlembac Adib Hasan
**Posts:**1016**Joined:**Tue Nov 22, 2011 7:49 pm**Location:**127.0.0.1-
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### Re: factorial

চমৎকার সমধান, ভাইয়া।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, $\boxed{n=8}$.

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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