MCQ
The remainder left out when ${8^{2n}} - {\left( {62} \right)^{2n + 1}}$ is divided by $9$ is :
- ✓$2$
- B$7$
- C$8$
- D$0$
$=(1+63)^{n}-(63-1)^{2 n+1}$
$=(1+63)^{n}+(1-63)^{2 n+1}$
$=\left(1+^{n} C_{1} 63+^{n} C_{2}(63)^{2}+\ldots+(63)^{n}\right.$
$+\left[1-^{(2 n+1)} C_{1} 63+(2 n+1) C_{2}(63)^{2}\right.$
$\left.+\ldots+(-1)(63)^{(2 n+1)}\right] $
$=2+63\left[^{n} C_{1}+^{n} C_{2}(63)+\ldots+(63)^{n-1}\right] $
$\left.+^{(2 n+1)} C_{2}(63)-\ldots-(63)^{(2 n)}\right]$
Hence, the remainder is $2.$
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