If x^y denotes x raised to the power y, find last two digits of (1941^3843) + (1961^4181)

• Concept 1: A number ending in 1
  
  Last two digits of (…a1)^(…b) will be [Last digit of a*b]1
  
  Ex 1.1 Last two digits of 1941^3843 = [Last digit of 4*3]1 = [Last digit of 12]1 = 21
  
  Ex 1.2 Last two digits of 1961^4181 = [Last digit of 6*1]1 = [Last digit of 6]1 = 61
  
  Ans is 21+61=82
• 5 years agoHelpfull: Yes(8) No(1)
• last digit in each number is 1,
  so for the first number last two digits:21
  last two digits for second number:61
  ans:82
• 5 years agoHelpfull: Yes(6) No(2)
• base last two digits ^power unit digit + base last two digits ^power unit digit=41^3 +61^1=21
• 5 years agoHelpfull: Yes(4) No(4)
• Multiply 4 from 1st number to 3 in the power, which will give 12. Taking 2 from 12 and 1 is already present at units place in 1941; hence this gives 21.
  Multiply 6 from 2nd number to 1 in the power, which will give 6. Taking 6 from 6 and 1 is already present at units place in 4181; hence this gives 61.
  Now 21+61=82. Hence the last 2 digits will be 82.
• 5 years agoHelpfull: Yes(1) No(0)
• 47 .
  find the relation of multiplication in a series format, the same two last 2 digits will return every time.
  the last 2 digits of 1st term should be 64 and of the second term is 83. hence it is 47.
• 5 years agoHelpfull: Yes(0) No(1)
• 41^3+61^1=21
• 5 years agoHelpfull: Yes(0) No(0)
• divide the power by 4 then put remainder as power and solve it now, multiple only 2 no. upto power
• 5 years agoHelpfull: Yes(0) No(0)