Dhaka Secondary 2009/7
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Please don't post problems (by starting a topic) in the "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.
Please don't post problems (by starting a topic) in the "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.
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 ![Smile :)](./images/smilies/icon_e_smile.gif)
![Smile :)](./images/smilies/icon_e_smile.gif)
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- 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.
$200^{th}$![Razz :P](./images/smilies/icon_razz.gif)
ain't this right.i am still working thinking this as the condition.
$200^{th}$
![Razz :P](./images/smilies/icon_razz.gif)
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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
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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
)
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
![Very Happy :D](./images/smilies/icon_e_biggrin.gif)
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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.