In this specific puzzle, it is $603!+604!+605!$.
If it was $603!*604!*605!$ instead, I would simply count all the occurrences of multiples of 5, 25 and 125, which would be $148$ for $603!$ and $604!$, and $149$ for $605!$, and add them together.
Yet in this puzzle the factorials are added together, so it could be either 148 zeroes—if the 149th zero of 605! gets overwritten—or more, if the digits which are not zero add up nicely, like in the solution, which says it's 150 zeroes at the end.
How can I determine if that will happen?