My train of thought here was to look for some sort of pattern as I add factorials and use that to deduce f(218)

1! is 1
2! is 2... 2! + 1! is 3
3! is 6... 3! + 2! + 1! is 9
4! is 24... 4! + 3! + 2! + 1! is 33
5! is 120... 5! + 4! + 3! + 2! + 1! is 153
6! is 720... We can stop right here since we can see a pattern where the ones digit is now 0.

Multiplying a number with a 0 at the ones digit is not going to change the ones digit; it will remain as a 0.
So even though we could add 8! and 9!, they will simply add some large (and annoying to calculate) number that has a 0 at the ones digit.

Ok, since we now know that 5!, 6!, 7!, ..., 218! are going to have a 0 at the ones digit, we know that adding those to 153, while yielding some large number, is simply adding 0 to the ones digit every time.
Thus, the ones will stay as 3 with no change. What is the remainder of a number with a ones digit of 3? It can only be 3 of course!!! (13%5 = 3, 14553%5 = 3, and 12314344983442342423423412348790213%3 = 3)

So the answer here is C

I hope this helps. This is the first question I answered on here.