15. $A, B, C$ are all positive integers. It is known that $A$ has 7 divisors, $B$ has 6 divisors, $C$ has 3 divisors, $A \times B$ has 24 divisors, $B \times C$ has 10 divisors. Then the minimum value of $A+B+C$ is $\qquad$ _
【Analysis】Since $\mathrm{A}$ has 7 divisors, we know that $A=p^{6}$, where $p$ is a prime number. Since $\mathrm{C}$ has 7 divisors, we know that $C=q^{2}$, where $q$ is a prime number.
If $B$ does not contain the prime factor $p$, then the number of prime factor $p$ in $A \times B$ is 6, and the number of divisors of $A \times B$ should be a multiple of $6+1=7$, which cannot be 24, so $B$ contains the prime factor $p$.
Since the divisors of 24 greater than 7 are $8, 12, 24$, the number of prime factor $p$ in $A \times B$ could be $7, 11, 23$.
If the number of prime factor $p$ in $A \times B$ is 11, then the number of prime factor $p$ in $B$ is 5. Since $B$ has 6 divisors, $B=p^{5}$. At this point, $B \times C=p^{5} \times q^{2}$. If $p=q$, then $B \times C=p^{7}$ has 8 divisors. If $p \neq q$, then $B \times C$ has 18 divisors, neither of which is 10 divisors, so the number of prime factor $p$ in $A \times B$ is not 11.
If the number of prime factor $p$ in $A \times B$ is 23, then the number of prime factor $p$ in $B$ is 17, which clearly cannot have 6 divisors.
Therefore, the number of prime factor $p$ in $A \times B$ is 7, and the number of prime factor $p$ in $B$ is 1. If $B$ does not contain the prime factor $q$, then the number of prime factor $q$ in $B \times C$ is 2, and the number of divisors of $B \times C$ should be a multiple of $2+1=3$, which cannot be 10, so $B$ contains the prime factor $q$.
Since the divisors of 10 greater than 3 are $5, 10$, the number of prime factor $q$ in $B \times C$ could be $4, 9$.
If the number of prime factor $q$ in $B \times C$ is 9, then the number of prime factor $q$ in $B$ is 7, which clearly cannot have 6 divisors.
Therefore, the number of prime factor $q$ in $B \times C$ is 4, and the number of prime factor $q$ in $B$ is 2.
So $B=p \times q^{2}$, and $p \neq q$.
Therefore, $A+B+C=p^{6}+p \times q^{2}+q^{2}$, the minimum value is $2^{6}+2 \times 3^{2}+3^{2}=91$.