find the last two digits of (1021^3921)+(3081^3921)

• when a no ends with 1 the last digit is 1.
  for last 2nd digit the short cut is
  1021-tenths place digit*unit place digit of the power=2(1)=2
  similarly for the second no 3081 it is 8(1)=8
  so the last two digits are 21+81=102.
  so answer is 02
• 11 years agoHelpfull: Yes(52) No(0)
• 02
  By using shortcut tricks
• 11 years agoHelpfull: Yes(24) No(1)
• 1021^3921=21
  3081^3921=81
  then 21+81=102
  so last two digit= 02
• 11 years agoHelpfull: Yes(10) No(1)
• Just Divide it by 100 to get the last two digits...
  (1021^3921 + 3081^3921)/100
  the term reduces to following by using successive division by 100 using remainder theroem.. 1^392*21+01^392*81 therefore 21+81 will give the remainder 2 when divided by 100..
• 11 years agoHelpfull: Yes(3) No(2)
• as we know that a^2+b^2=(a-b)^2+2ab
  so here a=3081 and b=1021
  on solving ans is 02
• 11 years agoHelpfull: Yes(2) No(2)
• @Krishna Chaitanya Gantasala
  can u plz tel me the short cut tricks...
• 11 years agoHelpfull: Yes(1) No(0)
• What are the tricks..??
  
• 11 years agoHelpfull: Yes(1) No(0)
• 01
  using shortcut tricks
• 11 years agoHelpfull: Yes(0) No(9)
• 01 is the answer as last 2 digits for 1021^3921 is 21 and for 3081^3921 is 81 thus 21+81 gives 01
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• 01 1^3921+1^3081
• 11 years agoHelpfull: Yes(0) No(2)
• 102
  1021^3921=(1020+1)^3921=(binomial expansion gives)=3921c0(1)^3921(1020)^0+3921c1(1)^3920(1020)^1.........(on solving)
  =(1*1+3921*1020)/100
  =1+20=21
  similarly 3081^3921 on solving gives 81
  so 21+81=102
• 11 years agoHelpfull: Yes(0) No(0)
• 2 as 1+1=2
• 11 years agoHelpfull: Yes(0) No(1)
• ansr is 02
  last digit of both the multiplication will be 1(1^nething=1)so adding them its 2.
  nd next in frst multiplication 2*2=4 nd in second 8*2=6(dropping out carry 1) so 4+6=0(dropping out carry)so
  02 is the answer :)
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• 02 are last two digits !!!
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• 5141^3921........
• 9 years agoHelpfull: Yes(0) No(0)