Question

The last digit in $7^{300}$ is

• A. 7
• B. 3
• C. 9
• D. 1

Answer 1

Answer: (D)

$7^{300}={\left(7^2\right)}^{150}=(50-1{)}^{150}$
$={}^{150}{C}_0(50{)}^{150}(-1{)}^0+{}^{150}{C}_1(50{)}^{149}(-1{)}^1$
$+\dots +{}^{150}{C}_{150}(50{)}^{\circ }(-1{)}^{150}$
Thus, the last digit of $7^{300}$ is ${}^{150}{C}_{150}\cdot 1\cdot 1=1$