Question

The last digit of the number ${\left(503\right)}^{2002}-{\left(501\right)}^{2001}$ is

• A. $1$
• B. $7$
• C. $8$
• D. $2$

Answer 1

The correct option is C $8$

$a={\left(503\right)}^{2002}-{\left(501\right)}^{2001}={(3+500)}^{2002}-{(1+500)}^{2001}$
$\Rightarrow a=(3^{2002}+{2002}_{C_1}{3}^{2001}{500}^1+...+{2002}_{C_{2002}}{500}^{2002})-(1+{2001}_{C_1}{500}^1+...+{2001}_{C_{2002}}{500}^{2002})$
$\Rightarrow a=3^{2002}-1+10\lambda ={(10-1)}^{1001}-1+10\lambda$
$\Rightarrow a={10}^{1001}-{1001}_{C_1}{10}^{1000}+{1001}_{C_2}{10}^{999}-.....+{1001}_{C_{1000}}10{-1001}_{C_{1001}}-1+10\lambda$
$\Rightarrow a=[10\lambda +100({10}^{999}-{1001}_{C_1}{10}^{998}+{1001}_{C_2}{10}^{997}-...-{1001}_{C_2})]+{1001}_{C_{1000}}10{-1001}_{C_{1001}}-1$
$\Rightarrow a=10k+1008=10k+1000+8=10l+8$
Therefore, unit place in $a$ is $8$.