10. (15 points) A natural number has exactly 9 distinct divisors, among which 3 divisors $A, B, C$ satisfy:
(1) $A+B+C=79$
(2) $A \times A=B \times C$
Then, this natural number is $\qquad$
【Solution】Solution: If a natural number $N$ has exactly 9 distinct divisors, then we can deduce that $N=x^{2} y^{2}$, or $N=x^{8}$. (1) When $N=x^{8}$, the nine divisors are: $1, x, x^{2}, x^{3}, x^{4}, x^{5}, x^{6}, x^{7}, x^{8}$, among which there are 3 divisors $A$, $B$, and $C$ that satisfy $A \times A=B \times C$, which is impossible.
(2) When $N=x^{2} y^{2}$, the nine divisors are: $1, x, y, x^{2}, x y, y^{2}, x^{2} y, x y^{2}, x^{2} y^{2}$, among which there are 3 divisors $A$, $B$, and $C$ that satisfy $A \times A=B \times C$,
(1) $A=x, B=1, C=x^{2}$, then $x+1+x^{2}=79$, no solution.
(2) $A=x y, B=1, C=x^{2} y^{2}$, then $x y+1+x^{2} y^{2}=79$, no solution.
(3) $A=x y, B=x, C=x y^{2}$, then $x y+x+x y^{2}=79$, no solution.
(4) $A=x y, B=x^{2}, C=y^{2}$, then $x y+x^{2}+y^{2}=79$, the solution is: $\left\{\begin{array}{l}x=3 \\ y=7\end{array}\right.$, then $N=3^{2} \times 7^{2}=441$.
(5) $A=x^{2} y, B=x^{2} y^{2}, C=x^{2}$, then $x^{2} y+x^{2} y^{2}+x^{2}=79$, no solution.

Therefore, the answer is 441.