just a simple logic-factorial-math counting

as you may know when dealing with factorial, such:
1! = 1
2! = 1 x 2
3! = 1 x 2 x 3
...sometimes we've met some interesting problems, like this one:

what number is in the last digits of ...
1! + 2! + 3! + 4! + 5! + ... + 208! + 209!

see the answer...

1! = 1
2! = 1 x 2
3! = 1 x 2 x 3
and for n! with n>4, you'll get
n! = 1 x 2 x 3 x 4 x 5 x ... x (n-1) x (n).
with 2 x 5 = 10
it means for all n>4, the last digit is always 0.
then to find the answer what we only need just 1! + 2! +3! + 4! = 1 + 2 + 6 + 24 = 33
so the answer is 3.
all the best for you.