Dhaka Higher Secondary 2011/9
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Please don't post problems (by starting a topic) in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Please don't post problems (by starting a topic) in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Dhaka Higher Secondary 2011/9
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 $f(1952) = 1$.
\[f(2^n)=f(2^{n+2})\text{ and } f\left (\sum_{n\in X}^{} 2^n\right)=\sum_{n\in X}^{} f(2^n)\]
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
Hint:
"Je le vois, mais je ne le crois pas!"  Georg Ferdinand Ludwig Philipp Cantor
Re: Dhaka Higher Secondary 2011/9
Is the first relation valid for $n =0$ and the second relation valid for all $X \subset \mathbb N\cup \{0\}$ ?
*edited
*edited
Every logical solution to a problem has its own beauty.
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Re: Dhaka Higher Secondary 2011/9
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.
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
Re: Dhaka Higher Secondary 2011/9
Solution:
Every logical solution to a problem has its own beauty.
(Important: Please make sure that you have read about the Rules, Posting Permissions and Forum Language)
(Important: Please make sure that you have read about the Rules, Posting Permissions and Forum Language)
Re: Dhaka Higher Secondary 2011/9
Yeah the camper is Mugdho and the answer was guessed during olympiad.
"Go down deep enough into anything and you will find mathematics." ~Dean Schlicter
Re: Dhaka Higher Secondary 2011/9
I mean he is one of them who wrote 0 in answer.
"Go down deep enough into anything and you will find mathematics." ~Dean Schlicter
Re: Dhaka Higher Secondary 2011/9
Mugdho did guess it and so did many others.
but I'm talking about someone solving it
but I'm talking about someone solving it
"Je le vois, mais je ne le crois pas!"  Georg Ferdinand Ludwig Philipp Cantor

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 Joined: Tue Dec 21, 2010 10:17 pm
Re: Dhaka Higher Secondary 2011/9
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.....
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.....
women of purity are for men of purity and hence men of purity are for women of purity  THE HOLY QURAN
Re: Dhaka Higher Secondary 2011/9
@the arrivals, check the question again
"Je le vois, mais je ne le crois pas!"  Georg Ferdinand Ludwig Philipp Cantor