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

A. 0

B. 1

C. 2

D. 3

Answer: Option D

Solution(By Examveda Team)

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,
$$\left(\overline{)1}+\overline{)2}+\overline{)3}+\overline{)4}\right)$$
$$\eqalign{ & = \left( {1 + 2 + 3 \times 2 \times 1 + 4 \times 3 \times 2 \times 1} \right) \cr & = \left( {1 + 2 + 6 + 24} \right) \cr & = 33 \cr} $$
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