Prime Divisor

For discussing Olympiad Level Number Theory problems
User avatar
Phlembac Adib Hasan
Posts:1016
Joined:Tue Nov 22, 2011 7:49 pm
Location:127.0.0.1
Contact:
Prime Divisor

Unread post by Phlembac Adib Hasan » Mon Feb 06, 2012 3:36 pm

Prove that for any prime $p\ge 7$, the number $(111...111)_{10}$ ,($(p-1)$ , $1$s) is a multiple of $p$.
Last edited by Phlembac Adib Hasan on Mon Feb 06, 2012 5:39 pm, edited 1 time in total.
Welcome to BdMO Online Forum. Check out Forum Guides & Rules

User avatar
nafistiham
Posts:829
Joined:Mon Oct 17, 2011 3:56 pm
Location:24.758613,90.400161
Contact:

Re: Prime Divisor

Unread post by nafistiham » Mon Feb 06, 2012 3:55 pm

Phlembac Adib Hasan wrote:Prove that for any prime $p\ge 7$, the number $(111...111)_{10}$ ,($(k-1)$ , $1$s) is a multiple of $p$.

could not understand it.what does $k$ depend on?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

User avatar
Phlembac Adib Hasan
Posts:1016
Joined:Tue Nov 22, 2011 7:49 pm
Location:127.0.0.1
Contact:

Re: Prime Divisor

Unread post by Phlembac Adib Hasan » Mon Feb 06, 2012 5:40 pm

nafistiham wrote:
Phlembac Adib Hasan wrote:Prove that for any prime $p\ge 7$, the number $(111...111)_{10}$ ,($(k-1)$ , $1$s) is a multiple of $p$.

could not understand it.what does $k$ depend on?
Oh,sorry :oops: It will be $p$.
Welcome to BdMO Online Forum. Check out Forum Guides & Rules

User avatar
Nadim Ul Abrar
Posts:244
Joined:Sat May 07, 2011 12:36 pm
Location:B.A.R.D , kotbari , Comilla

Re: Prime Divisor

Unread post by Nadim Ul Abrar » Mon Feb 06, 2012 9:33 pm

$(111....11)_{10} =\frac{10^{p-1}-1}{9}$

$(10,p)=1$
with fermat
$10^{p-1}-1\equiv 0 (modp)$

$p$ doesn't divide $9$
which leads
$\frac{10^{p-1}-1}{9}\equiv0 (modP)$
$\frac{1}{0}$

Post Reply