3. Find last two digits of : (1023^3921)+ (3081^3921)

• 23^3921+81^3921
  =23*23^3920+81^3921
  =23*(23^4)^980+(81)
  =23*(41)^980+81
  =23*01+81
  =104
  last 2 digit is 04(ans)
• 11 years agoHelpfull: Yes(18) No(7)
• 4 is the last digit
  
• 11 years agoHelpfull: Yes(5) No(7)
• for 1023 ..3921(last 21)=23
  
  for 3081.. 3921(last 21)=81
  
  23+81=104
  
  last digit 04(Ans.)
  
  
• 10 years agoHelpfull: Yes(1) No(0)
• 4 is the answer by using cyclicity
• 10 years agoHelpfull: Yes(0) No(0)
• 23+81=104
  last two digits are 04
• 10 years agoHelpfull: Yes(0) No(0)
• last two digit is =02
• 10 years agoHelpfull: Yes(0) No(8)
• 1023^3921 +3081^3921
  (1023^4)^980 *1023^3 + last two digits of 2nd exp are 81
  01*67+82=148
  Last rwo digit =48
  This is the only correct ans
• 10 years agoHelpfull: Yes(0) No(0)