$4 \cdot 30$ represents $E(n)$ as the largest integer $k$ such that $5^k$ is a divisor of the product $1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4 \cdots \cdots n^n$, then $E(150)$ equals what?
(Shanghai, China High School Mathematics Competition, 1988)
[Solution] Among $1,2,3, \cdots, 150$ there are $\left[\frac{150}{5}\right]=30$ multiples of 5. Removing one factor of 5 from each, along with their respective exponents, we get the exponent of 5 as
$$
\begin{aligned}
& 5+5 \cdot 2+5 \cdot 3+\cdots+5 \cdot 30 \\
= & 5(1+2+\cdots+30) \\
= & 2325 .
\end{aligned}
$$
Among $1,2, \cdots, 150$ there are $\left[\frac{150}{5^{2}}\right]=6$ multiples of $5^{2}$. Removing one factor of 5 from each, and then another, along with their respective exponents, we get the exponent of 5 as
$$
\begin{aligned}
& 5^{2}+5^{2} \cdot 2+\cdots+5^{2} \cdot 6 \\
= & 5^{2}(1+2+\cdots+6) \\
= & 525 .
\end{aligned}
$$
Among $1,2, \cdots, 150$, there is $\left[\frac{150}{5^{3}}\right]=1$ multiple of $5^{3}$. It has already had two factors of 5 removed, and we can remove one more, along with its respective exponent, to get the exponent of 5 as $5^{3}=125$. There are no multiples of $5^{k}(k \geqslant 4)$ among $1,2, \cdots, 150$, so
$$
E(150)=2325+525+125=2975 .
$$