A<B<C

From the hypotheses we get $9-A=\frac{A}{B} +\frac{B}{C} <\frac{B}{B} +\frac{B}{B}=2$ . This implies $A=8$ . By the equality we have: $\frac{8}{B}+\frac{B}{C}=1$ ; $8C+B^2=BC$ ; $C=B+8+\frac{64}{B-8}$ . Thus $B-8$ must be a divisor of $64$ , and because $B>A=8$ it must also be positive. Since we have $7$ positive divisors and $7$ corresponding values of $C$ satisfying the last equality the answer is $7$ .