Question

Find the remainder when $1!+2!+3!+4!+5!+...+49!$ is divided by $7$

• A. $0$
• B. $1$
• C. $5$
• D. $6$

Answer 1

Answer: (C)

$5$

When $1!+2!+3!+4!+5!......+49!$ is divided by $7$

Remainder $=$ Remainder of $1!+2!+3!+4!+5!+6!$

$=\frac{1+2+6+24+120+720}{7}$

$=\frac{873}{7}$

$=\frac{124}{7}+5$

$\Rightarrow$ Remainder $:5$

Hence, the correct answer is $5$.