AMC12 2004 12A, Problem 10

The sum of $49$ consecutive integers is $7^5$. What is their median?

Solution: Notice that $49\cdot7^3=7^5$. So, our middle number (median) must be $7^3\ $ 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 $\fbox{7^3}$.

Answer: 7^3