Dhaka Higher Secondary 2011/9

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Higher Secondary 2011/9

Unread post by BdMO » Fri Jan 28, 2011 10:32 pm

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 $f(1952) = -1$.

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Avik Roy
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Re: Dhaka Higher Secondary 2011/9

Unread post by Avik Roy » Fri Jan 28, 2011 11:35 pm

Hint:
Think binary
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Zzzz
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Re: Dhaka Higher Secondary 2011/9

Unread post by Zzzz » Sat Jan 29, 2011 9:46 am

Is the first relation valid for $n =0$ and the second relation valid for all $X \subset \mathbb N\cup \{0\}$ ?

*edited
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Avik Roy
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Re: Dhaka Higher Secondary 2011/9

Unread post by Avik Roy » Sat Jan 29, 2011 10:54 am

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.
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor

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Zzzz
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Re: Dhaka Higher Secondary 2011/9

Unread post by Zzzz » Sat Jan 29, 2011 11:44 am

Solution:
Base $2$ representation of $1952$ is $11110100000$
So, $1952=2^{10}+2^9+2^8+2^7+2^5$
$\Rightarrow f(1952)=f(2^{10}+2^9+2^8+2^7+2^5)$
$=f(2^{10})+f(2^9)+f(2^8)+f(2^7)+f(2^5)$

As, $f(2^n)=f(2^{n+2})$, $f(2^{10})=f(2^8)=f(2^6)=f(2^4)=f(2^2)=f(2^0)$
Similarly $f(2^9)=f(2^7)=f(2^5)=...=f(2^1)$

So, $f(1952)=2f(2^0) + 3f(2^1)=2f(1)+3f(2)$
$\therefore 2f(1)+3f(2)=-1... (1)$

Similarly $f(2011)= 5f(1)+4f(2)$
$\therefore 5f(1)+4f(2)=1... (2)$

Again $f(1971)=4f(1)+4f(2)... (3)$

From $(1)$ and $(2) \to f(1)=1,f(2)=-1$

From $(3) \to f(1971)=0$
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TIUrmi
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Re: Dhaka Higher Secondary 2011/9

Unread post by TIUrmi » Sun Jan 30, 2011 12:30 pm

Yeah the camper is Mugdho and the answer was guessed during olympiad.
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TIUrmi
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Re: Dhaka Higher Secondary 2011/9

Unread post by TIUrmi » Sun Jan 30, 2011 12:43 pm

I mean he is one of them who wrote 0 in answer.
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Avik Roy
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Re: Dhaka Higher Secondary 2011/9

Unread post by Avik Roy » Sun Jan 30, 2011 12:51 pm

Mugdho did guess it and so did many others.
but I'm talking about someone solving it ;)
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor

the arrivals
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Re: Dhaka Higher Secondary 2011/9

Unread post by the arrivals » Mon Jan 31, 2011 3:32 pm

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.....:(
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Avik Roy
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Re: Dhaka Higher Secondary 2011/9

Unread post by Avik Roy » Mon Jan 31, 2011 6:36 pm

@the arrivals, check the question again
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