Dhaka Higher Secondary 2011/8

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
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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.
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bristy1588
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Re: Dhaka Higher Secondary 2011/8

Unread post by bristy1588 » Wed Dec 14, 2011 1:29 pm

Can Aanyone tell me why i m wrong and what is the correct answer?
Bristy Sikder

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nafistiham
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Re: Dhaka Higher Secondary 2011/8

Unread post by nafistiham » Wed Dec 14, 2011 1:39 pm

হায় হায়! ফাহিম ভাইয়ারও এই উত্তর আসছে । আমি কয়েকজনকে তো এভাবেই দেখালাম । কিন্তু কিভাবে হবে ? এই টপিকে তো কোথাও সমাধানটা নাই ।
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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bristy1588
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Re: Dhaka Higher Secondary 2011/8

Unread post by bristy1588 » Wed Dec 14, 2011 1:48 pm

nafistiham wrote:হায় হায়! ফাহিম ভাইয়ারও এই উত্তর আসছে । আমি কয়েকজনকে তো এভাবেই দেখালাম । কিন্তু কিভাবে হবে ? এই টপিকে তো কোথাও সমাধানটা নাই ।
Tiham, Fahim er Solution ta ki??
Bristy Sikder

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Labib
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Re: Dhaka Higher Secondary 2011/8

Unread post by Labib » Thu Dec 15, 2011 1:08 am

I've got $6(115\cdot 146+11\cdot 147)$.
Really confused... :?
Please post your full solutions.... I'll post mine tomorrow.
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nafistiham
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Re: Dhaka Higher Secondary 2011/8

Unread post by nafistiham » Thu Dec 15, 2011 1:58 pm

unfortunately, my solution is so long that i hardly can even write it in my exercise book.so, i am trying to shorten it.whenever i can, i'll post
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
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nafistiham@gmail

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Labib
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Re: Dhaka Higher Secondary 2011/8

Unread post by Labib » Thu Dec 15, 2011 3:03 pm

Here's my proof... (short version)
There are $180$, '$3$' digit numbers divisible by $3$,having none of its digits same.(All the digits are non-zero)
Let me assign them to the set $S_1$.
Again there are $80$, '$3$' digit numbers divisible by $6$ and having none of its digits same.(All the digits are non-zero)
These $80$ are also the element of $S_1$.

Now, if the number $N= \overline{abcdefghi}$
Then $\overline{def}$ can be chosen in $80$ ways and there will be $179$ ways left to choose $\overline{abc}$.

For each of the $\overline{abcdef}$, $\overline{ghi}$ can be arranged in $6$ ways yielding the conclution that,
$N$ can have $6(80\cdot 179)$ different values.
Last edited by Labib on Thu Dec 15, 2011 3:52 pm, edited 1 time in total.
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Re: Dhaka Higher Secondary 2011/8

Unread post by Labib » Thu Dec 15, 2011 3:11 pm

Alright! done editing and the solution's ready to go...
Waiting for others to post their solution.
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mission264
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Re: Dhaka Higher Secondary 2011/8

Unread post by mission264 » Tue Jan 07, 2014 8:38 pm

i've got $N=2^6 \cdot 3^2 \cdot 37$

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Re: Dhaka Higher Secondary 2011/8

Unread post by Labib » Fri Jan 10, 2014 12:42 am

I think the solution should be $2^6.3^2.37$. But it doesn't seem to fit with the problem statement. :?
Ignore my previous stupid posts btw. :mrgreen:
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
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"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

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