Question

The remainder left out when $8^2-(62{)}^{2n+1}$ is divided by $9$ is

• A. $0$
• B. $2$
• C. $7$
• D. $8$

Answer 1

Answer: (B)

$2$

$8^2-(62{)}^{2n+1}$

$=(63+1)-(63-1{)}^{2n+1}$

$=(63+1)-{[}^{2n+1}{C}_0\left(63{)}^{2n+1}{+}^{2n+1}{C}_1\right(63{)}^{2n}(-1{)}^1+\dots \dots \dots \dots \dots .{+}^{2n+1}{C}_{2n}(63{)}^1(-1{)}^{2n}{}^{}{+}^{2n+1}{C}_{2n+1}(-1{)}^{2n+1}]$

Since $63$ is divisible by $9$, remainder $=1-\left[\right(-1{)}^{2n+1}]=2$

Thus B is correct answer