for n integer and p prime show that the exponent of the highest power of p appearing in the prime factorization of n! where
\[n=a_{k}p^k+a_{k-1}p^{k-1}+...a_{0}\]
is:
\[\frac{n-(a_{k}+...+a_{0})}{p-1}\]
prime factorization
- afif mansib ch
- Posts:85
- Joined:Fri Aug 05, 2011 8:16 pm
- Location:dhaka cantonment
- afif mansib ch
- Posts:85
- Joined:Fri Aug 05, 2011 8:16 pm
- Location:dhaka cantonment
Re: prime factorization
no1's solved yet?here's my one.
\[\left [ n/p \right ]+...\left [ n/p^k \right ]\]
\[=\frac{a_{1}(p-1)+...+a_{k}(p^k-1)}{p-1}\]
\[=\frac{n-(a_{1}+..+a_{k})}{p-1}\]
has any1 solved it in some other way????
\[\left [ n/p \right ]+...\left [ n/p^k \right ]\]
\[=\frac{a_{1}(p-1)+...+a_{k}(p^k-1)}{p-1}\]
\[=\frac{n-(a_{1}+..+a_{k})}{p-1}\]
has any1 solved it in some other way????
- nafistiham
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Re: prime factorization
afif mansib ch wrote,
maybe it should have been like this
\[\left [ n/p \right ]+...\left [ n/p^k \right ]\]
\[\left [ n/p \right ]+...\left [ n/p^k \right ]
maybe it should have been like this
\[\left [ n/p \right ]+...\left [ n/p^k \right ]\]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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.
- afif mansib ch
- Posts:85
- Joined:Fri Aug 05, 2011 8:16 pm
- Location:dhaka cantonment
Re: prime factorization
i wrote the same thing bt don't know why it came in such a form.
- nafistiham
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Re: prime factorization
actually, you missed two buttons.
i copied the line and added ' \] '
i copied the line and added ' \] '
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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.
- afif mansib ch
- Posts:85
- Joined:Fri Aug 05, 2011 8:16 pm
- Location:dhaka cantonment
Re: prime factorization
thanks for the correction via.