Official solution
\[{8^{2n}} - {\left( {62} \right)^{2n + 1}}\]
= \[{\left( {{8^2}} \right)^n} - {\left( {62} \right)^{2n + 1}}\]
= \[{\left( {1 + 63} \right)^n} - {\left( {1 - 63} \right)^{2n + 1}}\]
= \[\left( {1 + n.63 + {}^n{C_2}{{.63}^2} + ......} \right)\]
+ \[\left( {1 + {}^{2n + 1}{C_1}.\left( { - 63} \right) + {}^{2n + 1}{C_2}.{{\left( { - 63} \right)}^2} + ......} \right)\]
= 2 + 63\[\left[ {\left( {n + {}^n{C_2} + ....} \right) + \left( { - {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2}.63 + ......} \right)} \right]\]
= 63\[ \times \][Some integral value] + 2
63\[ \times \][Some integral value] + 2 by dividing with 9 we will get 2 as remainder as 63 is multiple of 9.