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The remainder left out when $8^{2 n}-(62)^{2 n+1}$ is divided by 9 is
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Verified Answer
The correct answer is:
2
2
$$
\begin{aligned}
& 8^{2 n}-(62)^{2 n+1}=(1+63)^n-(63-1)^{2 n+1} \\
& =(1+63)^n+(1-63)^{2 n+1}=\left(1+{ }^n c_1 63+{ }^n c_2(63)^2+\ldots .+(63)^n\right)+\left(1-{ }^{(2 n+1)} c_1 63+{ }^{(2 n+1)} c_2(63)^2+\ldots .+(-1)(63)^{(2 n+1)}\right) \\
& =2+63\left({ }^n c_1+{ }^n c_2(63)+\ldots .+(63)^{n-1}-{ }^{(2 n+1)} c_1+{ }^{(2 n+1)} c_2(63)+\ldots .(63)^{(2 n)}\right)
\end{aligned}
$$
$\therefore$ Reminder is 2
\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\right)+\left(1-{ }^{(2 n+1)} c_1 63+{ }^{(2 n+1)} c_2(63)^2+\ldots .+(-1)(63)^{(2 n+1)}\right) \\
& =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)}\right)
\end{aligned}
$$
$\therefore$ Reminder is 2
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