The question must be about divisors rather than prime factors, as the latter would allow the multiplication of 2013 by any set of 2010 primes other than 3, 11 and 61, the prime factors of 2013.

So, the numbers to be counted each must have 3, 11 and 61 as prime factors. In addition, each must have only 3 prime factors, occurring 2, 10 and 60 times respectively, so that it will have 2013 divisors.

That means that the 3! = 6 permutations of which repetition factor (i.e., power) goes with each of these prime factors accounts for all such multiples and the answer it 6.

The numbers are:

8235795981645330200581867774349676406483451904

 

7365509236983888976895651066110574636605059656318976

 

134671935428728472524601603815472075957729830026845044414383621408867483648

 

1895098522622937143400169618721365980441529983961299857326582053414894200392287473919

640300975603557084386466667167744

 

3935009126228450216653774822812473930016208009053604186319106962639822871333986893824

 

6191604918369242304773740220617917472927395579964585809115679830170324765117306520201

5833579707174402979500199993199820800

clc

p=perms([2,10,60]);

b=sym(2013);

f=factor(2013);

for i=1:length(p)

  disp(b*prod(f.^p(i,:)))

  disp(' ')

end