Question 938445
A number has more or less factors depending on its prime factorization.
If the prime factorization is
{{{n=A^a*B^b*"...."Y^y*Z^z}}}
those prime factors can be arranged into products of the form
{{{A^(alpha)*B^(beta)*"..."}}}
with {{{0<=alpha<=a}}} , {{{0<=beta<=b}}} , etc
to form all factors of {{{n}}}
from {{{1=A^0*B^0*"..."}}} to
{{{n=A^a*B^b*"..."}}} .
The number of possible products is
{{{(a+1)*(b+1)*"..."*(y+1)*(z+1)}}} ,
with as many such brackets as prime factors,
(multiplied together if there is more than one),
and each bracket being 2 or more.
In this case, {{{9=3*3}}} is the number of factors, 
meaning that there are just 2 prime factors,
both with 2 as an exponent:
{{{n=A^2*B^2}}} with {{{a=b=2}}} .
The smallest {{{n}}} we can make with that formula is
{{{n=2^2*3^2=4*9=36}}} .