8. Find the smallest integer \( n \) that has exactly 9 odd divisors and 18 even divisors, then the minimum value of \( n \) is
【Answer】 900
【Analysis】Let $n=2^{n} \times A$,
For the odd factors of $n$, the prime factor 2 cannot be chosen, while other prime factors can be chosen freely.
There are 1 multiplied by the number of factors of $A$ ways to choose, so the number of factors of $A$ is 9.
For the even factors of $n$, the prime factor 2 must be chosen, with 1 to $a$, giving $a$ ways to choose, and other prime factors have no restrictions, giving 9 ways.
Therefore, $a=2$.
$A$ has 9 factors, so $A=x^{8}$ or $A=x^{2} \times y^{2}$.
To find the smallest value of $n$, it is clearly $n=2^{2} \times A=2^{2} \times 3^{2} \times 5^{2}=900$.