16. Factorial $n$, written $n$ !, is defined by: $n!=1 \times 2 \times 3 \times \cdots \times n$. What is the remainder when $1!+2!+3!+4!+5!+6!+7!+8!+9!+10$ ! is divided by 5 ?
A 0
B 1
C 2
D 3
E 4
16. D For integer values of $n$ greater than $4, n$ ! is a multiple of 5 . So $5!+6!+7!+8!+9!+10$ ! is divisible by 5 .

Therefore the remainder when $1!+2!+3!+4!+5!+6!+7!+8!+9!+10$ ! is divided by 5 is the same as when $1!+2!+3!+4!$ is divided by 5 . Now $1!+2!+3!+4!=1+2+6+24=33=6 \times 5+3$. Therefore the required remainder is 3 .