Factorial sum

1! + 2! +3! +4! = 33 ≡ 3 (mod 10) ......... (i).
On the other side , 5! +6! + ..... +99! ≡ 0 (mod 10) [ as, each of 5!,6!, ... , 99! is divisible by 10] ........ (ii).
Adding, (i), (ii) we get, 1!+2! +........+ 99! ≡ 3 (mod 10). Hence the last digit of 1! +2! +.... +99! is 3, Which is the answer.

1!+2!+3!+4! = 1+2+6+24 = 33

After that, all factorial will finish with a 0 because its include the product of a 2 and 5 so the sum will always finish with a 3.

you need to consider only the first 4 terms 1! , 2! , 3! , 4! after that the last digit will be 0

so

1!+2!+3!+4!=33

the Last digit is 3

We know 5! ends with 0 so all no. above 5 have their factorials ending with 0 so now we left with 1!+2!+3!+4! =1+2+6+24 =33 So the result ends with 3...

From 5! on, the last digit of n! is 0. So the last digit of the given sum is the last digit of 1! + 2! + 3! + 4! = 33, so the last digit of the sum is 3.