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.

## Wednesday, September 24, 2008

### just a simple logic-factorial-math counting

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