55. Given that the sum of two positive integers is 432, the sum of the least common multiple and the greatest common divisor of these two positive integers is 7776. Then the product of these two positive integers is $\qquad$
Answer: 46620

Solution: Let the greatest common divisor of $a$ and $b$ be $d$, and $a=d m, b=d n$, then $m$ and $n$ are coprime, and the least common multiple of $a$ and $b$ is $d m n$. According to the conditions given in the problem,
$$
d m+d n=d(m+n)=432, \quad d m n+d=d(m n+1)=7776 .
$$

From the above two equations, we can get
$$
m n+1=18(m+n),
$$

which is
$$
m n-18(m+n)+1=0,
$$

thus
$$
(m-18)(n-18)=18^{2}-1=17 \times 19 \text {. }
$$

Assuming $m>n$, we have $m-18=17 \times 19=323, n-18=1$,

solving for
$$
\left\{\begin{array}{c}
m=341 \\
n=19
\end{array} ;\right.
$$

or $m-18=19, n-18=17$. Solving for $\left\{\begin{array}{l}m=37 \\ n=35\end{array}\right.$.
If $m=341, n=19$, then $m+n=360$, there does not exist a positive integer $d$ such that $d(m+n)=432$, so this solution is not valid.
If $m=37, n=35$, then $d=\frac{432}{37+35}=6$.

Therefore, $a=d m=222, b=d n=210, a$ and $b$'s product is 46620.