The Least Natural number $n$ which has $18$ divisors The Least Natural number $n$ which has $18$ divisors
My Try:: $18 = 2\times 3 \times 3 = 3^2 \times 2$
Now How can I solve it
Thanks

We know the number of divisors of $\prod_{1\le r\le m}p_i^{m_i}$ is $\prod_{1\le r\le m}(m_i+1)$ where $p_i$s are distinct primes, and power $m$s positive integers
As $18=2\cdot3^2$, $m$ can not be $\ge4$
If $m=1$, $m_1+1=18\implies m_1=17\implies n_\text{min}=2^{17}$
If $m=2$, $(m_1+1)(m_2+1)=18=6\cdot3=2\cdot9$
So, either $m_1=2$, $m_2=5\implies n_\text{min}=2^5 3^2=32\cdot9=288<2^{17}$ 
or  $m_1=1$,$m_2=8 \implies n_\text{min}=2^8 3=768$
If $m=3$, $m_1,m_2.m_3$ can only be $1,2,2$ so in this case, $n_\text{min}=2^2 3^2 5=180<288$
So, the minimum natural number with $18$ divisors is $180.$
