The sum of $49$ consecutive integers is $7^5$. What is their [median](https://artofproblemsolving.com/wiki/index.php/Median)?
$\text {(A)}\ 7 \qquad \text {(B)}\ 7^2\qquad \text {(C)}\ 7^3\qquad \text {(D)}\ 7^4\qquad \text {(E)}\ 7^5$
Solution 1
The median of a sequence is the middle number of the sequence when the sequence is arranged in order. Since the integers are consecutive, the median is also the [mean](https://artofproblemsolving.com/wiki/index.php/Mean), so the median is $\frac{7^5}{49} = 7^3\ \mathrm{(C)}$.

Solution 2
Notice that $49\cdot7^3=7^5$. So, our middle number (median) must be $7^3\ \mathrm{(C)}$ since all the other terms can be grouped to form an additional $48$ copies $7^3$. Adding them would give $7^5$.  
Solution by franzliszt