function

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tushar7
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function

Unread post by tushar7 » Sat Dec 18, 2010 11:46 pm

The function f, defined for whole positive numbers, is such that $f(1)=1$ and $f(2n) = 2f(n)$ and $f(2n+1)=4f(n) $for all$ n$. For how many numbers $n$ is$ f(n)=16$?

and please hide your solution . So any viewer can try it first.

AntiviruShahriar
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Re: function

Unread post by AntiviruShahriar » Sun Dec 19, 2010 2:12 am

$f(2n+1)=?$..........$f(2n)=2f(n)$ r $f(2n+1)$=kichu na hoile to ans may be 1......hidden kemne korree vai??????

HandaramTheGreat
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Re: function

Unread post by HandaramTheGreat » Sun Dec 19, 2010 9:54 am

AntiviruShahriar wrote:...hidden kemne korree vai??????
যেটা হাইড করতে চাও ঐটার প্রথমে লিখবা [ h i d e ]...(স্পেস বাদে)
আর শেষে লিখবা [ / h i d e ]...(স্পেস বাদে, স্পেস দিলাম যাতে এইটা নিজেই হাইড না হয়ে যায়)

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Labib
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Re: function

Unread post by Labib » Sun Dec 19, 2010 11:07 pm

I think the sol'ns are
$n=7,9,10,12,16$
.
Am I right?
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tushar7
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Re: function

Unread post by tushar7 » Mon Dec 20, 2010 12:14 am

yap ................ its correct . and did a good job at hiding too

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Labib
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Re: function

Unread post by Labib » Mon Dec 20, 2010 10:29 pm

thnx yaar!! :D
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.


"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

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Masum
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Re: function

Unread post by Masum » Tue Dec 21, 2010 8:27 pm

Above the edit box in a full view to reply,there are buttons like B,i,u,quote,hide etc.Press hide button and what you write inside gets hidden.
One one thing is neutral in the universe, that is $0$.

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