Question

The remainder when $27^{10}+7^{51}$ is divided by $10$

• A. 4
• B. 6
• C. 9
• D. 2

Answer 1

Answer: (D)

$(27{)}^{10}+7^{51}=(30-3{)}^{10}+(10-3{)}^{51}$
$=3^{10}-3^{51}+10\lambda$
$=(10-1{)}^5-3(10-1{)}^{25}+10\lambda$
$=-1+3+10{\lambda }_1$
$=2+10{\lambda }_1$
$\therefore$ Remainder is 2