**Problem 1:**

The sum of $81$ consecutive integers is $9^5$. What is their median?

Please

The sum of $81$ consecutive integers is $9^5$. What is their median?

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$(x-40)+(x-39)....+x+....(x+40)=9^5$

so, $81x=9^5$

so, $x=9^3$

median is $9^3$

so, $81x=9^5$

so, $x=9^3$

median is $9^3$

- Tahmid Hasan
**Posts:**665**Joined:**Thu Dec 09, 2010 5:34 pm**Location:**Khulna,Bangladesh.

Yup....it is $9^3=729$

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Nur Muhammad Shafiullah | Mahi

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

- nafistiham
**Posts:**829**Joined:**Mon Oct 17, 2011 3:56 pm**Location:**24.758613,90.400161-
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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.

Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.