Question

Let $N={\left(1001\right)}^3+{\left(1011\right)}^3+{\left(1000\right)}^3,$ the remainder when N is divided by 3! is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$0$

Let
$\left(1001\right)+\left(1011\right)+\left(1000\right)$
$=\left(3012\right)$
Dividing $3012$ by $3!$.
$=\frac{3012}{24}$
$=502$.
Or
$\left(1001\right)+\left(1011\right)+\left(1000\right)=6\left(502\right)+0$
Hence it will also divide $(1001{)}^3+(1011{)}^3+(1000{)}^3$ perfectly without leaving any remainder.