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
 A: 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 $180.$
A: The smallest prime power $n=p^k$ with 18 divisors would need $k=17$ and hence be $2^{17}$.
If $n=p^kq^m$ with two primes $p,q$ and $k\ge m\ge1$, then it has $(k+1)(m+1)$ divisors, which allows $k=8,m=1$ or $k=5,m=2$ and suggests $n=2^83$ or $n=2^53^2$. (Of course, it is best to take the smallest possible primes and use the biggest exponent with the smallest prime).
If $n=p^kq^mr^l$ correspondingly with $k\ge m\ge l\ge 1$, then we have $(k+1)(m+1)(l+1)$ divisors, which allows $k=m=2$, $l=1$ only and suggests $n=2^23^25$.
Four or more distinct prime divisors are not possible as that would produce either $2^4=16$ or at least $3\cdot 2^3=24$ divisors, but never $18$.
Choose the smallest number among $2^{17}$, $2^83$, $2^53^2$ and $2^23^25$.
A: $18=2\times 3\times 3$
So the power of the prime factors can be
case $1: 1,8$
case $2: 1,2,2$
case $3: 5,2$
case $4:17$
In the first case the smallest no. is $2^8\times3^1=256\times 3=768$
In the second case the smallest one is $5^12^23^2=36\times 5=180$
In the third case it is $3^22^5=9*32=288$
In the 4th case it is $2^{17}$
So clearly the smallest one is $180$.
Here I have used the fact that if $n=\prod_{i=1}^k p_i^{\alpha_i} $ then the no. of divisors of n are $\prod_{i=1}^k(\alpha_i+1)$
This can easily be verified as follows:
Any divisor of n can be of the form $\prod_{i=1}^k p_i^{\beta_i}$ where each $\beta _i$ can take values in $\{0,1,\dots,\alpha_i\}$ so there are $(\alpha_i+1)$ choices for each $\beta_i$.So the total no of such divisors =$\prod_{i=1}^k(\alpha_i+1)$
A: This isn't clear.  There are three possible questions in my opinion.  One is "What is the smallest natural number for which 18 is a divisor?"; the second is "What is the smallest natural number that has 18 divisors?"; and third, "What is the smallest natural number that has 18 distinct divisors?".  Can you clarify, please?
