2. 47 If $a<b<c<d<e$ are consecutive positive integers, $b+c+d$ is a perfect square, and $a+b+c+d+e$ is a perfect cube, what is the minimum value of $c$?
[Solution] Since $a, b, c, d, e$ are consecutive positive integers, we have
$$
\begin{array}{l}
b+c+d=3 c, \\
a+b+c+d+e=5 c .
\end{array}
$$

According to the problem, we can set
$$
\begin{array}{l}
3 c=m^{2} \quad(m \in N), \\
5 c=n^{3} \quad(n \in N),
\end{array}
$$

From (1), $m$ is a multiple of 3, thus $c$ is also a multiple of 3. From (2), $n$ is a multiple of 3, thus $c$ is a multiple of $3^{3}=27$. Also from (2), $n$ is a multiple of 5, thus $c$ is a multiple of 25. Therefore, $c$ is a multiple of $25 \times 27$, so $c \geqslant 25 \times 27$.

On the other hand, when $c=25 \times 27$, (1) and (2) can be satisfied, thus meeting the conditions of the problem. Therefore, the minimum value of $c$ is $25 \times 27=675$.