If $a<b<c<d<e$ are [consecutive](https://artofproblemsolving.com/wiki/index.php/Consecutive) [positive](https://artofproblemsolving.com/wiki/index.php/Positive) [integers](https://artofproblemsolving.com/wiki/index.php/Integer) such that $b+c+d$ is a [perfect square](https://artofproblemsolving.com/wiki/index.php/Perfect_square) and $a+b+c+d+e$ is a [perfect cube](https://artofproblemsolving.com/wiki/index.php/Perfect_cube), what is the smallest possible value of $c$?
Since the middle term of an [arithmetic progression](https://artofproblemsolving.com/wiki/index.php/Arithmetic_progression) with an odd number of terms is the average of the series, we know $b + c + d = 3c$ and $a + b + c + d + e = 5c$. Thus, $c$ must be in the form of $3 \cdot x^2$ based upon the first part and in the form of $5^2 \cdot y^3$ based upon the second part, with $x$ and $y$ denoting an [integers](https://artofproblemsolving.com/wiki/index.php/Integer). $c$ is minimized if it’s [prime factorization](https://artofproblemsolving.com/wiki/index.php/Prime_factorization) contains only $3,5$, and since there is a cubed term in $5^2 \cdot y^3$, $3^3$ must be a factor of $c$. $3^35^2 = \boxed{675}$, which works as the solution.