Find last two digits of (1021^3921)+ (3081^3921)

• unit place of both the no.s is 1
  10th place of 1021^3921=10th place of 1021*unit place of 3921=2*1=2
  10th place of 3081^3921=10th place of 3081*unit place of 3921=8*1=8
  so last two digits of (1021^3921)=21
  and last two digit of (3081^3921)=81
  so 21+81=102
  hence last two digits of (1021^3921)+ (3081^3921)=02
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• 1021^3921=21^3921=21
  3081^3921=81^3921=81
  last two digit=81+21=02..
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• 3081=980*4+1 so according to cyclicity 21+81=102..ans is 02
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• period if 21 and 81 is 5
  hence 21^3921=21 and 81^3921=81
  81+21=01
  hence 01 is answer
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• (3921/4)remainder=1
  so 1021^1+3081^1=4102
  so last two digit = 02
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• 02
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