## Dhaka Secondary 2009/7

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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### Dhaka Secondary 2009/7

Among the increasingly ordered permutations of the digits $1,2, \cdots ,7$ find the $2009^{th}$ integer.

amlansaha
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### Re: Dhaka Secondary 2009/7

in this case the 1st/lowest number is 1234567. so the 2009th number will be 1234567+2009-1=1236613 অম্লান সাহা

nafistiham
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### Re: Dhaka Secondary 2009/7

it is said that the permutations of $1,2,3,4,5,6,7$ which means the first number will be $1234567$ but, the second lowest number will be $1234576$ now we can see that the smallest number will be the number which has the biggest part same with $1234567$ in the left most order.
ain't this right.i am still working thinking this as the condition.

$200^{th}$ $\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
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amlansaha
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### Re: Dhaka Secondary 2009/7

how could i make such a silly mistake  অম্লান সাহা

nafistiham
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### Re: Dhaka Secondary 2009/7

firstly let us think of the smallest numbers . which is $1abcdef$.we have $720$ numbers like that.
thus, there are $720$ numbers like $2abcdef$,
$480$ numbers of these series $31abcde,32abcde,34abcde,35abcde$.
$72$ numbers of $361abcd,362abcd,364abcd$

then we have to see the $365abcd$ group. there are $24$ numbers like that the $17^{th}$ number is the answer.which is
$3654712$

i am not totally sure about the answer.but i think the way is right.please, post when you find any bug.

(starting the $3^{rd}$ century )
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
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