1.1941^3483 + 1961^4181= find out the last 2 digits
Ans: 82

• 1941^3483=> 1^3=1
  =>4*3=12ie 2
  hence last two digit of 1st part is 21
  lly 1961^4181=>1^1=1
  =>6*1=6
  hence last two digit=61
  therefore adding the above two parts 21+61
  we get 82
• 12 years agoHelpfull: Yes(12) No(6)
• 02
  bcoz 1^3=1, 4^3=64 leave 6, so 41
  1^1=1, 6^1=6 so 61
  41 + 61 = 102
• 12 years agoHelpfull: Yes(8) No(7)
• 1941^3483
  1^3=1 units place ,4*3=12 2 in tens place so 21
  1961^4181
  1^1=1 units place ,6*1 =6 6 in tens place 61 ..........21+61=82
• 12 years agoHelpfull: Yes(1) No(0)
• 1941^3483
  1^3=3 unit digit
  4^3=64 tens ---we take its unit digit 4
  now 41.
  simi.
  1961^4181
  61
  then 41+61=102
  
• 12 years agoHelpfull: Yes(1) No(0)