Question

The tens digits of $1!+2!+3!+...+49!$ is

• A. $3$
• B. $2$
• C. $16$
• D. $1$

Answer 1

The correct option is D $1$

We know that $1!+2!+3!+4!=33$

Also $5!=120,6!=720,7!=5040,8!=40320,9!=362880$

Thus ten's digit of $1!+2!+3!+4!...+9!$ is 1.

Also note that n! is divisible by $100$ for all $n\ge 10$

Thus the ten's digit of $1!+2!+3!+4!...+49!$ is $1$.