## Rangpur Secondary 2011/1

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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Moon
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### Rangpur Secondary 2011/1

Problem 1:
The sum of $81$ consecutive integers is $9^5$. What is their median?
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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 $9^3$
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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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Nur Muhammad Shafiullah | Mahi

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