Last digit problem

Note that since $5 \cdot 2 = 10$ and every multiple of $10$ has a units digit of zero, all factorials greater than $5!$ have a units digit of zero. Therefore, we only need to consider the first four factorials. Through simple arithmetic, one obtains that the units digit of $1!+2!+3!+4!$ is $3$ , so the answer is $\boxed{003}$ .

1!=1 , 2!=2 , 3!=6 , 4!=24 , 5!=120 , 6!=720...... and hence continue... So from here we can see that factorial of number greater than 4 will always end with zero. Hence unit digit of the sum of factorial till four (1!+2!+3!+4!=33) i.e. 3 will be at unit place 1!+2!+3!+4!+...+100!s as the unit place of number greater than 4! will be always zero.