9. (10 points) A positive integer, its double has exactly 2 more divisors than itself, and its triple has exactly 3 more divisors than itself. Then, this positive integer is $\qquad$
【Solution】Solution: This number can only contain 2 and 3 as prime factors, because if it had other prime factors, the final number of additions would be greater than the given number.

Let $x=2^{a} 3^{b}$, it has $(a+1)(b+1)$ divisors,
its double is $2^{a+1} 3^{b}$, which has $(a+1+1)(b+1)$ divisors,
then: $(a+1+1)(b+1)-(a+1)(b+1)=b+1=2$, solving for $b=1$;
Similarly, its triple is $2^{a} 3^{b+1}$, which has $(a+1)(b+1+1)$ divisors, which is 3 more than the original number,
i.e., $(a+1)(b+1+1)-(a+1)(b+1)=a+1=3$, solving for $a=2$,
so the form of this number is $2^{2} 3=12$;
Answer: This positive integer is 12.
Therefore, the answer is: 12.