Problem 1:
The sum of $81$ consecutive integers is $9^5$. What is their median?
Rangpur Secondary 2011/1
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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 $9^3$
so, $81x=9^5$
so, $x=9^3$
median is $9^3$
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Re: Rangpur Secondary 2011/1
Yup....it is $9^3=729$
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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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