Dhaka Secondary 2009/7

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BdMO
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Dhaka Secondary 2009/7

Unread post by BdMO » Fri Jan 21, 2011 6:18 pm

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

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amlansaha
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Re: Dhaka Secondary 2009/7

Unread post by amlansaha » Wed Nov 30, 2011 2:08 pm

in this case the 1st/lowest number is 1234567. so the 2009th number will be 1234567+2009-1=1236613 :)
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nafistiham
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Re: Dhaka Secondary 2009/7

Unread post by nafistiham » Wed Nov 30, 2011 11:53 pm

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.

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

Unread post by amlansaha » Thu Dec 01, 2011 11:24 am

how could i make such a silly mistake :oops: :oops:
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Re: Dhaka Secondary 2009/7

Unread post by nafistiham » Thu Dec 01, 2011 12:46 pm

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 :D )
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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