Question
The remainder left out when $${8^{2n}} - {\left( {62} \right)^{2n + 1}}$$ is divided by 9 is:
A.
2
B.
7
C.
8
D.
0
Answer :
2
Solution :
$$\eqalign{
& {8^{2n}} - {\left( {62} \right)^{2n + 1}} \cr
& = {\left( {64} \right)^n} - {\left( {62} \right)^{2n + 1}} = {\left( {63 + 1} \right)^n} - {\left( {63 - 1} \right)^{2n + 1}} \cr
& = \left[ {^n{C_0}{{\left( {63} \right)}^n} + {\,^n}{C_1}{{\left( {63} \right)}^{n - 1}} + {\,^n}{C_2}{{\left( {63} \right)}^{n - 2}} + ..... + {\,^n}{C_{n - 1}}\left( {63} \right) + {\,^n}{C_n}} \right] \cr
& = \left[ {^{2n + 1}{C_0}{{\left( {63} \right)}^{2n + 1}} - {\,^{2n + 1}}{C_1}{{\left( {63} \right)}^{2n}} + {\,^{2n + 1}}{C_2}{{\left( {63} \right)}^{2n - 1}} - ..... + \,{{\left( { - 1} \right)}^{2n + 1}}{\,^{2n + 1}}{C_{2 + 1}}} \right] \cr
& = 63 \times \left[ {^n{C_0}{{\left( {63} \right)}^{n - 1}} + {\,^n}{C_1}{{\left( {63} \right)}^{n - 2}} + {\,^n}{C_2}{{\left( {63} \right)}^{n - 3}} + .....} \right] + 1 - 63 \times \left[ {^{2n + 1}{C_0}{{\left( {63} \right)}^{2n}} - {\,^{2n + 1}}{C_1}{{\left( {63} \right)}^{2n - 1}} + .....} \right] + 1 \cr
& \Rightarrow \,\,63 \times {\text{some integral value}} + 2 \cr} $$
$$ \Rightarrow \,\,{8^{2n}} - {\left( {62} \right)^{2n + 1}}$$ when divided by 9 leaves 2 as the remainder.