minimum please!

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Rafe
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minimum please!

Unread post by Rafe » Thu Jan 17, 2013 10:02 pm

a,b,c,d,.....is a sequence.the number of factors of a,b,c,d,....is 2,3,4,5,....find the minimum value of a in natural number
Last edited by Rafe on Sat Jan 19, 2013 2:38 pm, edited 1 time in total.

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nafistiham
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Re: minimum please!

Unread post by nafistiham » Thu Jan 17, 2013 10:48 pm

Rafe wrote:$a,b,c,d,.....$ is a sequence.the number of products of $a,b,c,d,....$ is $2,3,4,5,....$ find the minimum value of a in natural number
How can a natural number have a limited number of products ?
Are you talking about factors ?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Prosenjit Basak
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Re: minimum please!

Unread post by Prosenjit Basak » Wed Jan 23, 2013 10:16 pm

Rafe wrote:a,b,c,d,.....is a sequence.the number of factors of a,b,c,d,....is 2,3,4,5,....find the minimum value of a in natural number
Factor means prime factor or only factor?
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Rafe
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Re: minimum please!

Unread post by Rafe » Thu Jan 24, 2013 5:27 pm

i mean all factors.such as 8 has 4 factors

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Fahim Shahriar
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Re: minimum please!

Unread post by Fahim Shahriar » Thu Jan 24, 2013 5:42 pm

I think you're talking about this sequence
$2,4,8,16.....,2^n$

$2^n$ has $(n+1)$ factors.
Name: Fahim Shahriar Shakkhor
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nafistiham
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Re: minimum please!

Unread post by nafistiham » Fri Jan 25, 2013 1:40 am

Fahim Shahriar wrote:I think you're talking about this sequence
$2,4,8,16.....,2^n$

$2^n$ has $(n+1)$ factors.
I think that is it. And, so, $a=2$

Prosenjit Basak
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Re: minimum please!

Unread post by Prosenjit Basak » Sat Jan 26, 2013 10:10 pm

I don't think so.You said $a=4$.But $4$ has $3$ factors. But in the question Rafe said $a$ has $2$ factors. So $a$ can't be $4$. I think Fahim Bhai may be correct.
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nafistiham
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Re: minimum please!

Unread post by nafistiham » Sun Jan 27, 2013 1:53 pm

Prosenjit Basak wrote:I don't think so.You said $a=4$.But $4$ has $3$ factors. But in the question Rafe said $a$ has $2$ factors. So $a$ can't be $4$. I think Fahim Bhai may be correct.
:oops: :oops: ...... that really was a lousy mistake. :oops:
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
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Prosenjit Basak
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Re: minimum please!

Unread post by Prosenjit Basak » Mon Jan 28, 2013 9:26 pm

Nothing to worry about that . To err is human........
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