Question 1087145
{{{16200=81*2*100=2^3*3^4*5^2}}} has many positive integral divisors,
from {{{1=1*1*1=2^0*3^0*5^0}}} to {{{16200=81*2*100=2^3*3^4*5^2}}} .
All of those divisors of the form
{{{2^a*3^b*5^c}}} ,
where {{{"a , b , and"}}}{{{c}}}
are integers such that
{{{system(0<=a<=3,0<=b<=4,0<=c<=2)}}} .
The ones that are odd must have {{{a+1=1}}} <--> {{{a=0}}} .
There are {{{1*(4+1)*(2+1)=1*5*3=15}}} of them.
The other {{{60-15=45}}} are even, and have {{{1<=a<=3}}} .
 
Each of those {{{2^a*3^b*5^c}}} divisors will have
{{{(a+1)*(b+1)*(c+1)}} positive integral divisors.
For that product to equal {{{4}}} we need one of those factors to be {{{1}}} .
The other two factors could be {{{4}}} and {{{1}}} , or {{{2}}} and {{{2}}} .
There is no other way.
 
For one of the factors to equal {{{4}}} , and the other two to equal {{{1}}} ,
it must be {{{a+1=4}}} <--> {{{a=3}}} ,
because {{{a+1=1}}} <--> {{{a=0}}} is not possible with an even divisor,
so the only way to do that is
{{{system(a=3,b+1=1,c+1=1)}}} <--> {{{system(a=3,b=0,c=0)}}} <--> {{{2^a*3^b*5^c=2^3*3^0*5^0=8*1*1=8}}} i.
That gives us {{{1}}} even divisors of {{{16200}}} with exactly four divisors.
 
All three factors can be equal to {{{2}}} :
{{{a+1=2}}} <--> {{{a=1}}} ,
{{{b+1=2}}} <--> {{{b=1}}} , and
{{{c+1=2}}} <--> {{{c=1}}} .
Choosing one of the factors than can equal {{{1}}} to be {{{1}}} , as in
{{{b+1=1}}} <--> {{{b=0}}} , or
{{{c+1=1}}} <--> {{{c=0}}} ,
and the other two factors to be {{{2}}} , 
we get {{{2}}} more even divisors of {{{16200}}} with exactly four divisors
{{{2^a*3^b*5^c=2^1*3^0*5^1=2*1*5=10}}} , and
{{{2^a*3^b*5^c=2^1*3^1*5^0=2*3*1=6}}} .
 
In sum, we can find {{{1+2=3}}} ways, and only {{{3}}} ways
to make an even divisor of {{{16200}}} with exactly four divisors.
The divisors of {{{16200}}} with exactly four divisors are
8, 10, and 6,
and they are only {{{3}}} of the {{{45}}}
positive integral divisors of {{{16200}}} that are even.
If a positive integral divisor of 16200 is selected of random,
and is one of the {{{45}}} that are even,
the chance that it will have exactly four divisors is
{{{3/45=highlight(1/15)}}}