The divisors of 2012 are 1, 2, 4, 503, 1006, 2012
Assuming that x, y and z don't have to be unique [ all different], we have
x y z
1 1 2012
1 2 1006
1 4 503
2 2 503
4 triplets ???
I think that's it [ could you check this, heureka???.....your always way better at this than I am....!!!]
