8. (10 points) There are three two-digit numbers $A$, $B$, and $C$. $A$ is a perfect square, and each of its digits is also a perfect square; $B$ is a prime number, and each of its digits is also a prime number, and the sum of its digits is also a prime number; $C$ is a composite number, and each of its digits is also a composite number, the difference between the two digits is also a composite number, and $C$ is between $A$ and $B$. What is the sum of $A$, $B$, and $C$? $\qquad$
【Solution】Solution: According to the analysis, first determine $A$,
$\because$ The one-digit numbers that are perfect squares are only $1,4,9$, and among them, the two-digit numbers that can form a perfect square are only $49, \therefore A=49$;
$\because$ The prime number $B$ has the sum of its two digits as a prime number and each digit is a prime number,
$\therefore$ The tens digit of $B$ can only be 2, and the units digit can only be 3, so $B=23$;
$\because$ The composite number $C$ has the difference between its two digits as a composite number and each digit is a composite number, then these digits can only be: $4,6,8,9$,
$C$ is between $A$ and $B$, i.e.,
$\therefore C=48$, hence $A+B+C=49+23+48=120$,
Therefore, the answer is: 120.