Answer:
210
Step-by-step explanation:
[tex]512=2^9 \\[/tex] has 9 prime factors
Now, if you want to maximize the number of different prime factor, apply the following method: let [tex]p(n)[/tex] be the nth prime number, in increasing order. Let [tex]v[/tex] define as follow : [tex]v(0)=1,v(n+1)=v(n)p(n+1)[/tex] , and find n such that [tex]v(n) < 1000,v(n+1) > 1000[/tex]. n will be the number of factors and v(n) your integer.
[tex]v(1)=2v(0)=2v(2)=3v(1)=6v(3)=5v(2)=30v(4)=7v(3)=210v(5)=11v(4)=2310[/tex]
There you are, 210 , with only 4 different prime factors.
→Credits: Xavier Dectot←
