For an integer n, n! = n(n - 1) (n - 2) ..... 3.2.1 Then, 1! + 2! + 3! +.....+ 100!, when divided by 5 leaves remainder

Answer: Option D

Every number from $$\overline{)5}$$ onwards is completely divisible by 5
$$\therefore \left(\overline{)5}+\overline{)6}+\overline{)7}+\mathrm{......}+\overline{)100}\right)$$       is completely divisible by 5
And,
$$\left(\overline{)1}+\overline{)2}+\overline{)3}+\overline{)4}\right)$$
$$\begin{array}{rl}& =\left(1+2+3\times 2\times 1+4\times 3\times 2\times 1\right)\\ & =\left(1+2+6+24\right)\\ & =33\end{array}$$
Clearly, 33 when divided by 5 leave a remainder 3
Hence,
$$\left(\overline{)1}+\overline{)2}+\overline{)3}+\overline{)4}+\overline{)5}+\mathrm{......}+\overline{)100}\right)$$        When divided by 5 leaves a remainder 3