Last two digits of the expression (1141^3843) + (1961^4181) ?

• Unit digit of this expression is always 1 as the base ends with 1. For the tenth place digit we need to multiply the digit in the tenth place of the base and unit digit of the power and take its unit digit.
  last two digits of (1141^3843)=(4*3),1=21
  last two digits of (1961^4181)=(6*1),1=61
  add 61+21
  so ans = 82
• 8 years agoHelpfull: Yes(10) No(0)
• 1141^3843=21
  1961^4181=61
  21+61=82
• 8 years agoHelpfull: Yes(3) No(1)
• we should use power cycle method
  take last 2 digit of 1141
  41^1=41 r=1
  41^2=81(take last 2 digits of ans) r=2
  41^3=21 r=3
  41^4=61 r=4
  41^5=01 r=0
  41^6=41
  hence power cycle is repeating and it is 5
  similarly for 1961
  61^1=61 r=1
  61^2=21 r=2
  61^3=81 r=3
  61^4=41 r=4
  61^5=01 r=0
  61^6=61
  hence for 1961 powercycle is 5
  for 1141 powercycle-5
  4181mod5=rem1=61
  for 1961 powercycle is 5
  3843mod5=rem 3=21
  therefore 21+61=82
• 8 years agoHelpfull: Yes(2) No(3)
• unit digit is 1^3 so 1 and and tens digit is 4^3 take that unit digit so 2 like other side
  ans is 82
• 8 years agoHelpfull: Yes(1) No(2)
• Unit digit of this expression is always 1 as the base ends with 1. For the tenth place digit we need to multiply the digit in the tenth place of the base and unit digit of the power and take its unit digit
  so ans = 82
  
• 8 years agoHelpfull: Yes(0) No(1)
• take last two digit of 1141 -> 4 x 3(unit digit of power) and (1^3 (unit digit of power)) = 21
  same for 1961 u get 61. then add both u get 82.
• 8 years agoHelpfull: Yes(0) No(1)