## Rangpur Secondary 2011/1

Problem for 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 "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.
Moon
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### Rangpur Secondary 2011/1

Problem 1:
The sum of $81$ consecutive integers is $9^5$. What is their median?
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

Hasib
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### Re: Rangpur Secondary 2011/1

$(x-40)+(x-39)....+x+....(x+40)=9^5$
so, $81x=9^5$
so, $x=9^3$

median is $9^3$
A man is not finished when he's defeated, he's finished when he quits.

Tahmid Hasan
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### Re: Rangpur Secondary 2011/1

i did the same way as hasib
বড় ভালবাসি তোমায়,মা

*Mahi*
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### Re: Rangpur Secondary 2011/1

Yup....it is $9^3=729$
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Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

nafistiham
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### Re: Rangpur Secondary 2011/1

It is true that, when $n=2k + 1$ the summation of any $n$ consecutive positive numbers is equal to $mn$ where $m$ is the median.
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.