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 $13$

$⟹X=1!+2!+3!+4!+5!+6!+7!+8!+9!+10!+........$
$⟹X=1+2+6+24+120+720+5040+40320+362780+3627800+........$
Now, starting from the $10!$, every factorial will have the last two digits $00$
i.e. the last two digits are given only by the sum of first $9$ terms
By adding the last two digits of the first nine terms we find that, the last two digits are $13$