PS: I can't understand the question
Candles....(From CMC test)
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always 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 "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always 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.
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Abdul Muntakim Rafi
- Posts:173
- Joined:Tue Mar 29, 2011 10:07 pm
- Location:bangladesh,the earth,milkyway,local group.
Bangla please...
PS: I can't understand the question
PS: I can't understand the question
Man himself is the master of his fate...
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bristy1588
- Posts:92
- Joined:Sun Jun 19, 2011 10:31 am
Re: Candles....(From CMC test)
Rafi,
Nowshin er kache $n$ sonkhok mombati ase. 1st din e, Noushin jekono 1 ta mombati 1 ghonta jalay, 2nd in e she jekono ekta mombati r prottekta 2 ghonta jalay, 3rd din e, she 3 ta mombati 3, prottektake 3 ghonta jalay,, evabe n tomo dine she n ta mombati n ghonta jalay, n-din sheshe, protita mombati shoman poriman jalano hoise. Tahole $n$ koto??
Nowshin er kache $n$ sonkhok mombati ase. 1st din e, Noushin jekono 1 ta mombati 1 ghonta jalay, 2nd in e she jekono ekta mombati r prottekta 2 ghonta jalay, 3rd din e, she 3 ta mombati 3, prottektake 3 ghonta jalay,, evabe n tomo dine she n ta mombati n ghonta jalay, n-din sheshe, protita mombati shoman poriman jalano hoise. Tahole $n$ koto??
Bristy Sikder
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Abdul Muntakim Rafi
- Posts:173
- Joined:Tue Mar 29, 2011 10:07 pm
- Location:bangladesh,the earth,milkyway,local group.
Re: Candles....(From CMC test)
Bristy,Thank you very much... 
Man himself is the master of his fate...
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nafistiham
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Re: Candles....(From CMC test)
excuse me, what is a CMC test.unfortunately, i don't know.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Re: Candles....(From CMC test)
Canadian Mathematics Contest. may be...
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
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bristy1588
- Posts:92
- Joined:Sun Jun 19, 2011 10:31 am
Re: Candles....(From CMC test)
Joty, Hehe he .. Its not Canadian Math Contest,, Its Chittagong Math Circle
Bristy Sikder
Re: Candles....(From CMC test)
Woww..... 
I'm a fool.
But there is actually a contest named Canadian Mathematics Contest.
amazing.....
I'm a fool.
But there is actually a contest named Canadian Mathematics Contest.
amazing.....
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
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bristy1588
- Posts:92
- Joined:Sun Jun 19, 2011 10:31 am
Re: Candles....(From CMC test)
এই সমস্যাটার কাছাকাছি একটা সমস্যা এবার Notre Dam এর অলিম্পিয়াডে আসছিল। পারি নাই। 
So no idia.
So no idia.
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
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Abdul Muntakim Rafi
- Posts:173
- Joined:Tue Mar 29, 2011 10:07 pm
- Location:bangladesh,the earth,milkyway,local group.
Re: Candles....(From CMC test)
In n days each candle will be lit for
\[\frac{(n+1)(2n+1)}{6}=k\]
Here k denotes the number of hours each candle would be lit which is obviously a positive integer...
That implies
\[2n^2+3n+1=6k\]
$6k$ is an even number... We can't get that if n is even... so let's assume n is odd... $n=2x+1$
Then we will get to the point
\[2(2x+1)^2+3(2x+1)+1=6k\]
\[2(x+1)(4x+3)=6k\]
\[(x+1)(4x+3)=3k\]
Now we need to prove this equation holds only for x=0 which implies n=1... Now the question is how do we prove this?
PS: I am just taking a guess it is possible only for n=1... Well if we can't prove this, we need to think otherwise..
\[\frac{(n+1)(2n+1)}{6}=k\]
Here k denotes the number of hours each candle would be lit which is obviously a positive integer...
That implies
\[2n^2+3n+1=6k\]
$6k$ is an even number... We can't get that if n is even... so let's assume n is odd... $n=2x+1$
Then we will get to the point
\[2(2x+1)^2+3(2x+1)+1=6k\]
\[2(x+1)(4x+3)=6k\]
\[(x+1)(4x+3)=3k\]
Now we need to prove this equation holds only for x=0 which implies n=1... Now the question is how do we prove this?
PS: I am just taking a guess it is possible only for n=1... Well if we can't prove this, we need to think otherwise..
Man himself is the master of his fate...