Question

The last two digits in $X=\sum _{k=1}^{100}k!$are?

• A. $10$
• B. $11$
• C. $12$
• D. $13$

Answer 1

The correct option is D

$13$

Explanation for the correct option:

$$\begin{gathered} X=1!+2!+3!+4!+5!+6!+7!+8!+9!+10!+... \\ X=1+2+6+24+120+720+5040+40320+362780+3627800+... \end{gathered}$$

After$10!$, we will have$00$ as the last two digits for all the factorials.

Add the sum of the last two digits of $1$to $9$ factorials

$1+2+6+4+0=13$

The last two digits of $X=\sum _{k=1}^{100}k!$is $13$

Hence, option D is correct.