<s>[INST] 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
 [/INST] 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 $2^23^25$.
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