If x^y denotes x raised to the power y, find last two digits of (1941 ^ 3843)+ (1961 ^4181).
a. 02 b. 82 c. 42 d. 22

• Step 1: (1941^3843)
  Take last two digit 41 from 1941 and raise it to the unit digit power 3 from 3843. (i.e) 41^3 = 68921. From this take only the last two digit 21.
  Step 2: (1961^4181)
  Take last two digit 61 from 1961 and raise it to the unit digit power 1 from 4181. (i.e) 61^1=61.
  Step 3:
  Add that 21 + 61 = 82
• 9 years agoHelpfull: Yes(40) No(1)
• we should use power cycle method
  take last 2 digit of 1941 take last 2 digits of 1961
  
  41^1=41 r=1 61^1=61 r=1
  41^2=81(take last 2 digits of ans)r=2 61^2=21 r=2
  41^3=21 r=3 61^3=81 r=3
  41^4=61 r=4 61^4=41 r=4
  41^5=01 r=0 61^5=01 r=0
  41^6=41 61^6=61
  hence power cycle is repeating and it is 5 hence for 1961 it is 5
  r-remainder for 1961 powercycle-5
  for 1941 powercycle-5 4181mod5=rem1=61
  3843mod5=rem 3=21
  therefore 21+61=82
• 12 years agoHelpfull: Yes(31) No(2)
• Remember 1 raised to any power will give 1 as unit digit.
  To find the digit in the 10th place, we have to multiply, 10th digit in the base and unit digit in the power.
  So the Last two digits of the given expression = 21 + 61 = 82
• 10 years agoHelpfull: Yes(16) No(4)
• we should use power cycle method
  take last 2 digit of 1941
  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 1941 powercycle-5
  4181mod5=rem1=61
  for 1961 powercycle is 5
  3843mod5=rem 3=21
  therefore 21+61=82
  
  
• 12 years agoHelpfull: Yes(14) No(0)
• last two digits of 1941^3843 is 21
  last two digits of 1961^4181 is 61
  so the last two digits of the whole sum is 82
• 12 years agoHelpfull: Yes(13) No(30)
• 1941^3843
  it is a rule that numbers ending with 1 has at the end 1 at unit place as it is and at tens place we get as given below
  multiply tens place of base number(1941) which is 4 and unit place of power(3843) which is 3
  so 4*3=12 so consider 2 only
  and here 21
  similarly second one1961^4181 have 81
  when they add up 21+81=01 ans
  
• 9 years agoHelpfull: Yes(7) No(2)
• Remember 1 raised to any power will give 1 as unit digit.
  To find the digit in the 10th place, we have to multiply, 10th digit in the base x unit digit in the power.
  
  So the Last two digits of the given expression = 21 + 61 = 82
  
• 10 years agoHelpfull: Yes(5) No(0)
• Can you explain last two digits of 1941^3843 is 21?? how.?
• 10 years agoHelpfull: Yes(5) No(0)
• 81+21=102
  ans02
• 12 years agoHelpfull: Yes(3) No(18)
• ANS IS : 82
  Remember 1 raised to any power will give 1 as unit digit.
  To find the digit in the 10th place, we have to multiply, 10th digit in the base and unit digit in the power.
  So the Last two digits of the given expression = 21 + 61 = 82
• 9 years agoHelpfull: Yes(3) No(0)
• spanda way of think that problem and getting answer is i hope correct ... thank"S spandana
  
• 12 years agoHelpfull: Yes(2) No(3)
• last two digits of 1941^3843 is 21
  last two digits of 1961^4181 is 61
  so the last two digits of the whole sum is 82
• 10 years agoHelpfull: Yes(1) No(1)
• 3843 should in the form of [4n+x], therefore x=3. so find last 2 digits of [1941]^3, we get 21.
  similarly for 4181., we get x=1, so [1961]^1,we get 61.
  Therefore laste two digits is 21+61=82
• 9 years agoHelpfull: Yes(1) No(0)
• explain clearly@spandana
• 12 years agoHelpfull: Yes(0) No(5)
• 82
  
  
  
  
  
  
  
  
• 9 years agoHelpfull: Yes(0) No(1)
• Remember 1 raised to any power will give 1 as unit digit.
  To find the digit in the 10th place, we have to multiply, 10th digit in the base x unit digit in the power.
  
  
  So the Last two digits of the given expression = 21 + 61 = 82
• 9 years agoHelpfull: Yes(0) No(0)
• Here Answer is 82
• 8 years agoHelpfull: Yes(0) No(0)
• 1+1=2 last digit
  4+6=10 therefore 0 is second last digit
  
  And : 02
• 6 years agoHelpfull: Yes(0) No(0)