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

Question

Aicharya · Accepted Answer

$ \begin{aligned} & 8^{2 n}-(62)^{2 n+1}=(1+63)^n-(63-1)^{2 n+1} \ & =(1+63)^n+(1-63)^{2 n+1}=\left(1+{ }^n c_1 63+{ }^n c_2(63)^2+\ldots .+(63)^night)+\left(1-{ }^{(2 n+1)} c_1 63+{ }^{(2 n+1)} c_2(63)^2+\ldots .+(-1)(63)^{(2 n+1)}ight) \ & =2+63\left({ }^n c_1+{ }^n c_2(63)+\ldots .+(63)^{n-1}-{ }^{(2 n+1)} c_1+{ }^{(2 n+1)} c_2(63)+\ldots .(63)^{(2 n)}ight) \end{aligned} $ $	herefore$ Reminder is 2

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

Solution