Rangpur Secondary 2011/1

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Moon
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Rangpur Secondary 2011/1

Unread post by Moon » Wed Feb 02, 2011 6:45 pm

Problem 1:
The sum of $81$ consecutive integers is $9^5$. What is their median?
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Re: Rangpur Secondary 2011/1

Unread post by Hasib » Wed Feb 02, 2011 6:50 pm

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

median is $9^3$
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Re: Rangpur Secondary 2011/1

Unread post by Tahmid Hasan » Wed Feb 02, 2011 7:19 pm

i did the same way as hasib :D
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Re: Rangpur Secondary 2011/1

Unread post by *Mahi* » Thu Feb 03, 2011 8:20 pm

Yup....it is $9^3=729$
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Re: Rangpur Secondary 2011/1

Unread post by nafistiham » Fri Jan 06, 2012 6:06 pm

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