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Dhaka Higher Secondary 2011/9
Posted: Fri Jan 28, 2011 10:32 pm
by BdMO
Consider a function $f: \mathbb N$ $\to$ $\mathbb Z$ is so defined that the following relations hold:
\[f(2^n)=f(2^{n+2})\text{ and } f\left (\sum_{n\in X}^{} 2^n\right)=\sum_{n\in X}^{} f(2^n)\]
where $X$ is some finite subset of $\mathbb{N} \cup \{0\}$.
Find $f(1971)$ if it is known that $f(2011) = 1$ and $f(1952) = -1$.
Re: Dhaka Higher Secondary 2011/9
Posted: Fri Jan 28, 2011 11:35 pm
by Avik Roy
Re: Dhaka Higher Secondary 2011/9
Posted: Sat Jan 29, 2011 9:46 am
by Zzzz
Is the first relation valid for $n =0$ and the second relation valid for all $X \subset \mathbb N\cup \{0\}$ ?
*edited
Re: Dhaka Higher Secondary 2011/9
Posted: Sat Jan 29, 2011 10:54 am
by Avik Roy
Yes, $n=0$ is allowed in the first relation as well
btw, i guess it should be mentioned that only one camper could actually 'solve' this problem during the olympiad and he found the perfect solution.
Re: Dhaka Higher Secondary 2011/9
Posted: Sat Jan 29, 2011 11:44 am
by Zzzz
Re: Dhaka Higher Secondary 2011/9
Posted: Sun Jan 30, 2011 12:30 pm
by TIUrmi
Yeah the camper is Mugdho and the answer was guessed during olympiad.
Re: Dhaka Higher Secondary 2011/9
Posted: Sun Jan 30, 2011 12:43 pm
by TIUrmi
I mean he is one of them who wrote 0 in answer.
Re: Dhaka Higher Secondary 2011/9
Posted: Sun Jan 30, 2011 12:51 pm
by Avik Roy
Mugdho did guess it and so did many others.
but I'm talking about someone solving it
Re: Dhaka Higher Secondary 2011/9
Posted: Mon Jan 31, 2011 3:32 pm
by the arrivals
whats problem with this solution??
1971=S+32+16+2+1
1995=S+64+8+2+1
2011=S+64+16+8+2+1 where S=1024+512+256+128
as f(2^n)=f(2^[n+2])
set f(S)=H(S)
so f(1971)=H(S)+2(f(1)+f(2))
and f(1995)=H(S)+2(f(1)+f(2))
so f(1995)=f(1971)=-1.....
Re: Dhaka Higher Secondary 2011/9
Posted: Mon Jan 31, 2011 6:36 pm
by Avik Roy
@the arrivals, check the question again