Q47 / 90 · JEE Main 2009 · Mathematics

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

  • A
    0
  • Correct answer
    B
    2
  • C
    7
  • D
    8

Solution

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

Hence, the correct answer is option: 'B'.

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