14. A number is the product of three prime factors. The squares of these three prime factors; the sum is 2331, the number 7560 is less than this number and is coprime with it, and the sum of its divisors (excluding 1 itself) is 10560. Find this number.
Let $c, b, c$ be the prime factors of a number, then
$$
a^{2}+b^{2}+c^{2}=2331.
$$

In addition, we know that the number of integers less than this number and coprime to it is
$$
\begin{array}{l}
a b c\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\left(1-\frac{1}{c}\right). \\
\text { That is, }(a-1)(b-1)(c-1)=7560.
\end{array}
$$

According to the problem, we also have
$$
(a+1)(b+1)(c+1)=10560.
$$

From (2) and (3), we get:
$$
a b c+a+b+c=9060.
$$

And $b c+c a+a b+1=1500$.
From (1) and (5), we get
$$
\begin{array}{l}
(a+b+c)^{2}=5329. \\
\text { Hence } a+b+c=73.
\end{array}
$$

From (4), we get
$$
a b c=8987=11 \cdot 19 \cdot 43.
$$

The solution to this problem uses two theorems in number theory: the first, the number of positive integers less than a given number and coprime to it, if $N$ represents this number, and $N=a^{p} b^{q} c^{r} \cdots$, where $a, b, c, \cdots$ are different prime numbers, and $p, q, r, \cdots$ are positive integers, then $\phi(N)=N\left(1-\frac{1}{a}\right) \cdot\left(1-\frac{1}{b}\right) \cdot\left(1-\frac{1}{c}\right) \cdots$. The second, "the number of divisors of a number", if $N$ represents this number, and $N=a^{p} b^{q} c^{r} \cdots$, where $a, b, c, \cdots$ are different prime numbers, and $p, q, r, \cdots$ are positive integers, then the number of divisors is $(p+1)(q+1)(r+1) \cdots$.