Verified Solution • 2025–26 Session moderate

QJEEM22Ma6cb6: JEE Main 2022 Mathematics - Class 11 JEEM22Ma6cb6 Solution: The remainder when is divided by 9 is

The remainder when (11)1011+(1011)11(11)^{1011}+(1011)^{11} is divided by 9 is

Multiple Choice Options

  • A 1
  • B 4
  • C 6
  • D 8

Step-by-Step Verified Solution

Re((11)1011+(1011)119)=Re(21011+3119){\mathop{\rm Re}\nolimits} \left( {{{{{(11)}^{1011}} + {{(1011)}^{11}}} \over 9}} \right) = {\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}} + {3^{11}}} \over 9}} \right)

For Re(210119){\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}}} \over 9}} \right)

21011=(9−1)337=337C09337(−1)0+337C19336(−1)1+337C29335(−1)2+  .....  +  337C33790(−1)337{2^{1011}} = {(9 - 1)^{337}} = {}^{337}{C_0}{9^{337}}{( - 1)^0} + {}^{337}{C_1}{9^{336}}{( - 1)^1} + {}^{337}{C_2}{9^{335}}{( - 1)^2} + \,\,.....\,\, + \,\,{}^{337}{C_{337}}{9^0}{( - 1)^{337}}

So, remainder is 8

and Re(3119)=0{\mathop{\rm Re}\nolimits} \left( {{{{3^{11}}} \over 9}} \right) = 0

So, remainder is 8

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