10. Let $m$ be the product of all positive integer divisors of 360,000 . Suppose the prime factors of $m$ are $p_{1}, p_{2}, \ldots, p_{k}$, for some positive integer $k$, and $m=p_{1}^{e_{1}} p_{2}^{e_{2}} \cdot \ldots p_{k}^{e_{k}}$, for some positive integers $e_{1}, e_{2}, \ldots, e_{k}$. Find $e_{1}+e_{2}+\ldots+e_{k}$.
Answer: 630
Solution: Let $d(m)$ is the number of postive divisors of $m$. Since $360000=2^{6} \cdot 3^{2} \cdot 5^{4}$, we have $m=360000 \frac{d(m)}{2}$. Thus,
$$
e_{1}+e_{2}+e_{3}=\frac{(6+2+4)}{2} \cdot[(6+1)(2+1)(4+1)]=630 .
$$