We decompose a number into its prime factors; let $N=a^{\alpha} \times b^{\beta} \times c^{\gamma}$. First, we need to list all divisors of this number; second, determine the number of all divisors; third, prove that if these divisors are arranged in increasing order, the product of two divisors equidistant from the ends is equal to $N$; finally, calculate the product of all divisors of the given number. The results obtained should be applied to the number 630.
$1^{0}$. Every divisor of $N$ is of the form

$$
a^{m} b^{n} c^{p}
$$

where the exponents $m, n$, and $p$ can be equal to zero and are at most equal to $\alpha, \beta$, and $\gamma$ respectively.

To write down all divisors of $N$, it suffices to assign every value from 0 to $\alpha$ to $m$, every value from 0 to $\beta$ to $n$, and every value from 0 to $\gamma$ to $p$. In other words, the divisors in question are given by the terms of the following product:

$$
\left(1+a+a^{2}+\cdots+a^{\alpha}\right)\left(1+b+b^{2}+\cdots+b^{\beta}\right)\left(1+c+c^{2}+\cdots+c^{\gamma}\right)
$$

$2^{0}$. The number of divisors is equal to the number of terms in the above product, which is

$$
(\alpha+1)(\beta+1)(\gamma+1)
$$

$3^{0}$. Let $d$ and $d_{1}$ be two divisors that are equidistant from the ends. I claim that

$$
d=\frac{N}{d_{1}}
$$

By assumption, there are an equal number of divisors of $N$ between 1 and $d$ and between $d_{1}$ and $N$. If I divide $N$ by the sequence of divisors from $d_{1}$ to $N$, I obtain the sequence of divisors of $N$ from $\frac{N}{d_{1}}$ to $\frac{N}{N}=1$. But these, when arranged in ascending order, start with the same number and have the same number of terms as the sequence from 1 to $d$, and thus $\frac{N}{d_{1}}$ must be equal to $d$, that is,

$$
d d_{1}=N
$$

$4^{0}$. It follows from the above that every divisor of $N$ is of the form

$$
\frac{N}{d_{1}}
$$

where $d_{1}$ is some divisor of $N$.

If we multiply all the divisors together, the product takes the following form:

$$
P=\frac{N^{(\alpha+1)(\beta+1)(\gamma+1)}}{P}
$$

or

$$
P=\frac{N^{(\alpha+1)(\beta+1)(\gamma+1)}}{2}
$$

Application. If $N=630=2 \times 3^{2} \times 5 \times 7$, the divisors of $N$ are then $1,2,3,5,7,10,14,15,18,21,30,35,42,45,63,70,90,1$ and 630; their number is

$$
(1+1)(2+1)(1+1)(1+1)=24
$$

and their product is

$$
630^{12}=3909188328478827879681 \cdot 10^{12}
$$