Actually, it will have to be a multiple of 9, because 50! is a multiple of 9, and the sum of the digits of any multiple of 9 is itself a multiple of 9. This is more or less easy to show. As for finding the actual sum of the digits of 50! by analytic means, I'm still not sure how I'd do it.
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So 50! = 30414093201713378043612608166064768844377641568960512000000000000 and the sum is....216. But I can't get there other than by brute force.
Praise the speed of modern computers! Yeah I was stumped too. I know there is probably a clever way, but I was stuck just with "brute force" as well
Yes, it's very simple to compute that in many programming languages (I use Python) or tools (like Mathematica). I'm not even sure if there is an efficient analytical way to do it. I have play around a bit with it, but so far to no avail.