Advanced P17(BOMC 2)

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*Mahi*
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Advanced P17(BOMC 2)

Unread post by *Mahi* » Sun Apr 01, 2012 7:21 pm

Prove that the sequence $1, 11, 111, \cdots $ contains an infinite subsequence whose terms are pairwise relatively prime.
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Re: Advanced P17(BOMC 2)

Unread post by *Mahi* » Mon Apr 02, 2012 2:49 am

Hint:
$\gcd(a^x-1,a^y-1)=a^{\gcd(x,y)}-1$
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Nadim Ul Abrar
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Re: Advanced P17(BOMC 2)

Unread post by Nadim Ul Abrar » Mon Apr 02, 2012 8:19 pm

*Mahi* wrote:Hint:
$\gcd(a^x-1,a^y-1)=a^{\gcd(x,y)}-1$
Then we can consider the terms of form $\frac {10^p-1}{9}$ ; where $p$ is prime .
we have infinity many primes ,SO we can find infinity many pairwise relatively prime terms .
$\frac{1}{0}$

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Re: Advanced P17(BOMC 2)

Unread post by Phlembac Adib Hasan » Tue Apr 03, 2012 1:11 pm

This came to my head about two years ago.I also solved like Nadim Vaia.It's the easiest way,I think.
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Re: Advanced P17(BOMC 2)

Unread post by zadid xcalibured » Tue Apr 03, 2012 1:36 pm

i solved like mahi's.it's the smartest way.

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Re: Advanced P17(BOMC 2)

Unread post by nafistiham » Tue Apr 03, 2012 1:57 pm

Pardon me, what is a sub-sequence ?
Must it have a general form for each of it's terms ?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Advanced P17(BOMC 2)

Unread post by *Mahi* » Tue Apr 03, 2012 2:33 pm

Sub-sequence means a sequence formed by only the elements of another elements in the same order.
(It is quite similar to set and subset, eliminate a few elements from the first and you will find the other.)
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