Question

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

• A. 7
• B. 8
• C. 0
• D. 2

Answer 1

Answer: (D)

$8^{2n} - 62^{2n+1} = (a - 1)^{2n} - (63 - 1)^{2n+1}$
$= (^{2n}c_{0}9^{2n}-\,{}^{2n}c_{1}9^{2n-1}+\,{}^{2n}c_{2}9^{2n-2} .....$
$- \,{}^{2n}c_{2n-1}\,\left(9\right)+\,{}^{2n}c_{2n})$
$- \,{}^{2n+1}c_{0}\left(63\right)^{2n+1}-\,{}^{2n+1}c_{1}\left(63\right)^{2n}$
$+\,{}^{2n+1}c_{2}\,\left(63\right)^{2n-1} ...........$
$+\,{}^{2n+1}c_{1}\left(63\right)-\,{}^{2n+1}c_{0}$
$= 9_{m+1+1} = 9_{m+2}$ for some integer $m$.
Thus $8^{2n} - 62^{2n+1}$ is divided by $9$, the remainder is $2$.