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

• last 2 digit of 1021^3921 is 21
  last 2 digit of 3081^3921 is 81
  21+81=102
  thus,last 2 digit is=02
  
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• 1021^3921 + 3081^3921=21+81=02
  last digit =02
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• last 2 digit are 01
  as 1021^3921=last digit 21
  and 3081^3921 =last digit 81
  add both last digit will be 102====so last two digit is 02
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• last 2 digit= 02
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• last 2 digit are 01
  as 1021^3921=last digit 41
  and 3081^3921 =last digit 61
  add both last digit will be 01
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• answer is 02
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• in 1021^3961, last 2 digits are 21..similarly in 3081^3921 =81
  21+81=102
  hence last 2 digits are 02
  
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• last 2 digits of 1021^3921 is 21
  last 2 digits of 3081^3921 is 81
  therefore, 21+81=102
  so the last 2 digit is = 02
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• 2^1 + 8^1
  =2 + 8
  that is ,21+81=102
  so,02 is ans
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• period if 21 and 81 is 5
  hence 21^3921=21 and 81^3921=81
  81+21=02
  hence 02 is answer
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• 2 any power of will give the lasr digit 1 then add
  
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• 1^x=1
  x=any integer
  1021^3921 last digits are 21
  3081^3921 last digits are 81 (tenth place digit 8 multiplied with unit digit in power)
  so ans=102
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