5. (10 points) A positive integer $A$ when divided by 3! results in a number whose number of factors is $\frac{1}{3}$ of the original number of factors. What is the smallest $A$ that satisfies this condition? $\qquad$
【Solution】Solution: Assume the prime factorization form of the original number is $2^{a} \times 3^{b}$, the number of factors is $(a+1)(b+1)$, and the result after multiplying by 3 ! is $2^{a-1} \times 3^{b-1}$. According to the problem, $(a+1)(b+1)=3 a b$. To make $A$ as small as possible, $b$ should be as small as possible. When $b=1$, it is easy to find that $a=2$. Therefore, the smallest $A$ is $2^{2} \times 3=12$. Hence, the answer is 12.