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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
Hint:
Think binary

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
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 $(3) \to f(1971)=0$

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