## Dhaka Secondary 2009/7

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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.

### 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
**Posts:**829**Joined:**Mon Oct 17, 2011 3:56 pm**Location:**24.758613,90.400161-
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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.

$200^{th}$

ain't this right.i am still working thinking this as the condition.

$200^{th}$

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

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.

- nafistiham
**Posts:**829**Joined:**Mon Oct 17, 2011 3:56 pm**Location:**24.758613,90.400161-
**Contact:**

### 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 )

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 )

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

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.