Question

The remainder obtained when $1!+2!+......50!$ is divided by $30$ is

• A. $3$
• B. $13$
• C. $11$
• D. $9$

Answer 1

The correct option is A $3$

Consider $i^{th}$ term : $i!$.
For $i>4,t_i=t_5\ast \left(6\right)\left(7\right)....(i-1)\left(i\right)$ i.e. a multiple of $t_5$.
$t_5=5!=120$ is a multiple of 30. Hence, $t_i(i>4)$ are multiples of 30 and leave reminder of 0.
Answer $=$ $(1!+2!+3!+4!)mod\left(30\right)=33mod\left(30\right)=3$
(here, a mod (b) is the reminder of left by a when divided by b)