@rz923 That's an interesting thought! If you'd encountered this problem on a math competition and, say, didn't have enough time to think of the way Prof. Loh demonstrated in the video, I'd say this would work!
It's not completely rigorous, though. I definitely agree with you that you can find the first 3 digits by doing 369 + 36 + 3 + (whatever # is carried into the 3rd column), but how do you know for sure that that carried # is 2? For example, if you actually were to write out the whole sum of 369369369's, you'd see that the carried digit, from right to left, starts as 1, but goes all the way up to 5, and then down to the 2 you mentioned. It is a very reasonable guess that the 3rd column carried digit is 2-- it definitely cannot be one because the 4th column has at least 3+6+9+3 = 21. And the only way that the digit carried into the 3rd column could be 3 is if the digit carried into the 4th column is 9. This isn't really likely.
Basically, the way you proposed is definitely valid, but if you wanted to be 100% sure, you could also use the other method.