Verified Solution • 2025–26 Session moderate

QJEEM09Maee22: JEE Main 2009 Mathematics - Class 11 JEEM09Maee22 Solution: The remainder left out when is divided by 9 is :

The remainder left out when 82n−(62)2n+1{8^{2n}} - {\left( {62} \right)^{2n + 1}} is divided by 9 is :

Multiple Choice Options

  • A 2
  • B 7
  • C 8
  • D 0

Step-by-Step Verified Solution

82n−(62)2n+1{8^{2n}} - {\left( {62} \right)^{2n + 1}}

= (82)n−(62)2n+1{\left( {{8^2}} \right)^n} - {\left( {62} \right)^{2n + 1}}

= (1+63)n−(1−63)2n+1{\left( {1 + 63} \right)^n} - {\left( {1 - 63} \right)^{2n + 1}}

= (1+n.63+nC2.632+......)\left( {1 + n.63 + {}^n{C_2}{{.63}^2} + ......} \right)
         + (1+2n+1C1.(−63)+2n+1C2.(−63)2+......)\left( {1 + {}^{2n + 1}{C_1}.\left( { - 63} \right) + {}^{2n + 1}{C_2}.{{\left( { - 63} \right)}^2} + ......} \right)

= 2 + 63 [(n+nC2+....)+(−2n+1C1+2n+1C2.63+......)]\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.

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