Example 15 (1989 7th American Invitational Mathematics Examination) 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 $c$?
Since $a, b, c, d, e$ are consecutive positive integers, we have
$$
\begin{array}{l}
b+c+d=3 c, \\
a+b+c+d+e=5 c .
\end{array}
$$

Since $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube, we can set
$3 c=m^{2}$,
$5 c=n^{3}$.
From (1), we get $3 \mid m$, and thus $3 \mid c$.
From (2), we get $5 \mid n$, and thus $5^{2} \mid c$.
From (2) again, we get $3^{3} \mid c$.
Therefore, $25 \cdot 27 \mid c$.
Since $c \geqslant 25 \cdot 27=675$,
the minimum value of $c$ is 675.