jmc

We use the fact that the number of divisors of a number $n=p_1^{e_1}{p}_2^{e_2}\cdots p_k^{e_k}$ is $(e_1+1)(e_2+1)\cdots (e_k+1)$. If a number has $18=2\cdot 3\cdot 3$ factors, then it can have at most $3$ distinct primes in its factorization. Dividing the greatest power of $2$ from $n$, we have an odd integer with six positive divisors, which indicates that it either is ($6=2\cdot 3$) a prime raised to the $5$th power, or two primes, one of which is squared. The smallest example of the former is $3^5=243$, while the smallest example of the latter is $3^2\cdot 5=45$. Suppose we now divide all of the odd factors from $n$; then we require a power of $2$ with $\frac{18}{6}=3$ factors, namely $2^{3-1}=4$. Thus, our answer is $2^2\cdot 3^2\cdot 5=\boxed{180}$.