Find last two digits of : (1021^3921)+ (3087^3921)

• for (1021^3921) => last two digit = 21
  unit digit = 1, tenth = 10th place digit of 1021 * unit of 3921 = 2*1 = 2
  
  for (3087^3921) = (87^4)^980 *87 = (------61)^980 * 87 = 01 * 87 = 87
  
  last two digits of (1021^3921)+ (3087^3921) = 21 + 87 = 108 => 08
  
  
• 9 years agoHelpfull: Yes(23) No(0)
• 21+87=108
  i.e Ans. 08
• 9 years agoHelpfull: Yes(5) No(0)
• use technique of x^4n+1
  consider unit digit of both terms : 1^1 + 7^1 = 08
• 9 years agoHelpfull: Yes(3) No(0)
• please explain it again
  
• 9 years agoHelpfull: Yes(1) No(0)
• 21+87=108
  i.e 08
• 9 years agoHelpfull: Yes(1) No(0)
• 41+67=128
  i.e 28
• 9 years agoHelpfull: Yes(0) No(3)
• ans:63
  21,41,61,81,01 for 21
  87,69,03,61,07,01 for 87
  21*03=63
• 9 years agoHelpfull: Yes(0) No(1)
• take unit digit of the no. and add them and take last two digits of ans
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• ans 02
  21+81 =102
• 9 years agoHelpfull: Yes(0) No(0)
• go through this tutorial http://www.pagalguy.com/articles/winning-silver-figuring-out-the-second-last-digit-in-quantit-8784711
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