Question

The last digit of $(1!+2!+.......+2005!{)}^{500}$ is

• A. $9$
• B. $2$
• C. $7$
• D. $1$

Answer 1

The correct option is D $1$

$S=(1!+2!+3!+4!+5!+....2005!)$

The last digit of $(5!)$ onwards every term will be zero.

$\Rightarrow S=(1+2+6+24+120+720+....2005!)$

$=(33+10\lambda );$ $\lambda \in I$

Hence, the unit digit of $S$ is $3.$

$\Rightarrow S=33+10\lambda =3+30+10\lambda =(3+10K);$ $K\in I$

$\Rightarrow S^{500}=({3+10K)}^{500}=(3^{500}{+}^{500}{C}_1.3^{499}10K+....)$

$=(3^{500}+10{\lambda }_0);$ ${\lambda }_0\in I$

$\Rightarrow 3^{500}=9^{250}=(10-1{)}^{250}$

$=({10}^{250}{-}^{250}{C}_1.{10}^{249}+......+(1{)}^{250})$

$=10{\lambda }_1+1;$ ${\lambda }_1\in I$

$\Rightarrow$ Hence, last digit of $S^{500}=1.$

Hence, the answer is $1.$